course 201
10/27
Class 091021short-answer questions
`q001. ** Give your answers to the following one-step questions:
What is the KE of a 10 kg object moving at 5 m/s?
What is `dPE for a 10 kg object raised 3 meters against the force of gravity?
What is `dPE for a block of dominoes pulled 15 cm in such a way to stretch a spring, if as a result the spring force changes linearly from 4 N to 16 N?
If on a certain interval, the KE and PE of a system change so that `dKE = + 8 J and `dPE = -12 J, then how much work is done by nonconservative forces on the system?
If on a certain interval nonconservative forces do 20 J of work on a system and its KE increases by 15 J, does the PE of the system increase or decrease, and by how much?
****
What is the KE of a 10 kg object moving at 5 m/s?
KE=.5mv^2=.5(10 kg)(5 m/s^2)=5 kg(25 m^2/s^2)=125 kgm^2/s^2=125 J
What is `dPE for a 10 kg object raised 3 meters against the force of gravity?
`dPE=-WorkONbyGravity=-Fgrav*`ds=-(m*a)*`ds=-(10 kg*9.8 m/s^2)*3 meters=-98 kgm/sec^2*3 m= 294 kgm^2/sec^2=294 J
What is `dPE for a block of dominoes pulled 15 cm in such a way to stretch a spring, if as a result the spring force changes linearly from 4 N to 16 N?
Since there is no change in vertical height, the effect of gravity does not change, and thus the potential energy of the domino block does not change.
If on a certain interval, the KE and PE of a system change so that `dKE = + 8 J and `dPE = -12 J, then how much work is done by nonconservative forces on the system?
`dWncON=`dPE+`dKE=-12 J+8 J=-4 J
If on a certain interval nonconservative forces do 20 J of work on a system and its KE increases by 15 J, does the PE of the system increase or decrease, and by how much?
`dWncON=`dPE+`dKE, 20 J=`dPE + 15 J, `dPE=5 J, so the potential energy increased by 5 Joules
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`q002. ** A slingshot shoots a pebble in the upward direction. While the slingshot is in contact with the pebble:
Does the elastic force exerted by the slingshot act in the direction of the pebble's motion or opposite this direction?
Does the elastic force therefore do positive or negative work on the pebble?
Does its elastic PE therefore increase or decrease?
Does the force of gravity act in the direction of the pebble's motion or opposite this direction?
Does the gravitational force therefore do positive or negative work on the pebble?
Does its gravitational PE therefore increase or decrease?
Assuming that the gravitational and elastic forces are the only conservative forces acting on the pebble, does its total PE increase or decrease?
****
The elastic force acts in the direction of the pebbles motion. I know this because the rubber of the slingshot travels forward with the pebble, but is stopped from completing its motion due to its attachment to the slingshot system.
The elastic force therefore does positive work on the pebble
The elastic PE increases. This can be proven as tension was put into the rubber band as the slingshot was pulled back before release.
The force of gravity works in the opposite direction of the pebbles outward motion, pulling it to the ground
The gravitational force therefore does negative work on the pebble
Gravitational PE decreases as the pebble starts from a high position and falls to the ground
The total potential energy of the pebble decreases completely. At first it had elastic potential energy behind it with the stretched rubber band as well as gravitational PE as it was higher than the ground. But after being projected forward, it lost contact with the rubber band and lost its elastic potential energy, and after being pulled all the way to the ground its gravitational potential energy was also reduced to zero.
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`q003. ** For the ball rolling in two directions along the 'constant-velocity' incline (i.e., an incline on which the velocity of the ball as it rolls down the incline remains unchanged), as you observed it in class:
What were your data?
Explain how you determined the acceleration of the ball the roll in each direction.
If the ramp was completely level, would it be a 'constant-velocity' incline?
If the ramp was completely level, what would be the acceleration in each direction?
On a 'constant-velocity' ramp, how are the parallel component of the gravitational force and the force of rolling friction related? (the 'parallel component' of the gravitational force is the component of that force in the direction of the incline)
****
The 61 cm ramp was nearly level, and a ball rolling from left to right had constant velocity.
A ball rolling from right to left, however, did not attain constant velocity.
It came to a stop at 44.5 cm in 10.4 sec, at 48.5 cm in 10.9 sec, at 53 cm in 11.6 seconds.
In the right direction, there was no acceleration as there was no change in velocity. A=`dv/`dt, and if `dv=0, a=0. In the left direction, acceleration can be found by using Vave=`ds/`dt, then v0=2*Vave (because vf=0), then a=`dv/`dt. I arrived at accelerations of .8 cm/sec^2, .82 cm/sec^2, and .79 cm/sec^2.
If the ramp was completely level, it would be considered a constant-velocity incline, but in reality the ball would need an initial push to get it started, creating an higher initial velocity, and it would slow to a stop by the end of the incline. The constant velocity would refer to the middle portion of the incline only.
If the ramp was completely level, the acceleration in each direction would be zero. This would be due to the zero change in velocity that I described above.
The positive work done by gravity in the X direction (level with the incline) would be equal and opposite to the negative work done by friction, slowing the ball down as it rolls.
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`q004. The rolling friction between a steel ball and a steel incline exerts a force opposed to the direction of motion. If this force is equal to .3% of the mass of the rolling ball, then
What is the frictional force on a 60 gram steel ball?
What would be the acceleration of the ball on a perfectly level ramp?
What would be its acceleration down a ramp with a 1 degree slope?
What would be its acceleration up the same ramp?
****
Ffriction=-.003(Fnormal)=-.003(.06 kg*9.8 m/sec^2)=-.003(.59 N)=-.002 Newtons
Acceleration on a perfectly level ramp would be zero
Let alpha=1 degree. MgX for this ramp would be .59 N*cos(1)=.59N.
you need to measure the angle of the weight from the positive x axis. That angle would be 269 deg or 271 deg, depending on how you draw your picture. It should be clear from your drawing that the x component is very small compared to the magnitude of the weight.
Fnet=Ffriction +MgX=-.002 N+.59N=.588 N. also, fnet=ma, so .588N=m*a=.06 kg*a, so a=9.8 m/sec^2
Up the same ramp, the ball would be traveling in the opposite direction, working in the same direction as the force of friction, so both forces would be considered negative. Fnet=Ffriction +MgX=-.002 N+-.59N=-.59 N. also, fnet=ma, so -.59N=m*a=.06 kg*a, so a=-9.9 m/sec^2
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`q005. The ball in the preceding rolls 50 cm down the 1 degree ramp, subject to only gravity, normal force and rolling friction. For this roll:
What is `dPE?
What is `dW_NC_ON?
What therefore is `dKE?
If the ball started from rest, what therefore is its final velocity?
****
`dPE=-WbyGravOn=-Fgrav*`ds=-(.59N)*.5 m=-.3N
the ball doesn't roll .5 meters in the direction of the gravitational force; rather .5 m in the direction of the x component of that force
I am unsure of how to solve for `dWncON without the `dWnet
`dW_NC is the result of rolling friction; you calculated the force of rolling friction previously so you can get `dW_NC_ON.
If I had `dWncON, I would find `dKE either by using `dKE=-`dPE+`dWncON
I am unsure of how to solve for final velocity here.
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`q006. The acceleration of a 60-gram steel ball on a certain ramp with a small incline is observed to be 2.5 cm/s^2 when the ball moves in one direction, and -1.5 cm/s^2 when the ball moves in the opposite direction.
What is the net force on the ball as it moves in each direction?
Assuming that the force of rolling friction has the same magnitude in both directions, what is the force of rolling friction on the ball?
Is the ramp slopes in the positive or the negative direction?
What is the slope of the ramp?
****
One direction: fnet=ma=60 g (2.5 cm/sec^2)=150 gcm/sec^2, opposite direction: fnet=60 g(-1.5 cm/sec^2)=-90 gcm/sec^2
If rolling friction is the same in both directions, then the difference in forces in each direction are not due to friction, they are instead due to the ramp. This means ..?
Not sure how to calculate this
see the given solution on this one; briefly friction accounts for the difference, and friction acts in opposite directions when the velocities are opposite, so the difference in the calculated forces is double the force of friction.
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`q007. ** My mass is about 75 kg. In about 40 minutes, my altitude and velocity change as follows:
I start at rest on a level runway in Atlanta, accelerate to 50 m/s before beginning to rise, rise to an altitude of 10 000 meters at a velocity of 200 m/s, descend to a level runway 300 miles away which I first contact at a speed of 60 m/s, and come to rest at the other end of the runway.
This scenario naturally breaks into five intervals, one occurring on the first runway, another between the end of that runway and cruising altitude, a third between cruising altitude and the beginning of descent, the fourth between the beginning of the descent and first contact with the second runway, and the last on the second runway.
Give quantitative answers to each of the following:
What happens to my gravitational potential energy during each phase?
What happens to my kinetic energy during each phase?
What nonconservative forces act during each phase?
What net work is done by nonconservative forces during each phase?
****
Gravitational PE does not change during phases 1, 3, or 5, at which the plane is moving parallel to the ground, not changing its altitude at all. In phase 2, gravitational PE increases as the plane works against gravity to obtain altitude. In Phase 4, gravitational PE decreases as the plane descends towards the ground, working with the force of gravity.
During Phase 1, kinetic energy increases as velocity increases. This can be explained by the direct relationship between KE and velocity in the equation KE= .5mv^2. This is also true of phase 2, at which velocity increases significantly. During phase 3 velocity is kept relatively constant, so there is no significant change in kinetic energy, but kinetic energy is still very high because the velocity of the plain is still very high. During phase 4 kinetic energy decreases as the plane slows, catching wind resistance as it falls through the air. During phase 5 kinetic energy returns to zero as the velocity of the plane comes back to zero.
In phase 1, air resistance and the turbines of the plane serve as nonconservative forces. The turbines are a positive force, while the air resistance is a negative force. The same is true of phases two and three and four. During phase five, air resistance plays a bigger part in slowing the plane, and the brakes of the plane work as a non conservative force as well, slowing the plane to a stop.
In phase one, the turbines do a significant amount of work while air resistance does little work against the plane as it is aerodynamic. The same is true of stage two and three. In phase 4 air resistance does much more work as the pilot positions the plane so it becomes less aerodynamic, therefore allowing it to slow and descend towards the runway. In phase 5 air resistance and the planes brakes do a significant amount of work in slowing the plane.
Very good. See the given solutions for some additional details about the nature of the nonconservative forces.
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`q008. A certain Atwood machine has 50 paperclips on each side. The paperclips have reasonably uniform mass. A positive direction has been declared, and the following observations are made.
If one of the clips is transferred to the positive side the system remains at rest.
If another clip is transferred to the positive side the system still remains at rest.
If a third clip is transferred to the positive side, the system accelerates at 15 cm/s^2.
After restoring the system to its original state, the system remains at rest after a transfer of 2 clips to the negative side.
A third clip transferred to the negative side results in an acceleration of - 10 cm/s^2.
First answer the following:
If there was no friction involved, what acceleration would result from transferring a single paperclip, so that one side has 51 clips and the other 49 clips?
****
It seems to me that it would matter which side of the system the paperclip was added to.
The sign of the acceleration would depend on the side; the magnitude would not. See the given solution.
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Now, in terms of the stated data:
As a multiple of the weight of a paperclip, what do you conclude is the magnitude of the frictional force resisting the motion of the system?
****
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`q009. ** A 40 kg block rests on an incline which makes an angle of 37 degrees with the horizontal. The coefficient of friction between the block and the incline is mu = .5. Sketch a free-body diagram for this block, sketch x and y axes with the x axis parallel to the ramp, and estimate the x and y components of the weight of the block. Then using the sine and cosine functions, calculate these components.
Give your estimates of the x and y components, as well as the calculated x and y components of the weight.
Using the fact that the block remains in equilibrium in the y direction, determine the normal force between block and incline.
What therefore is the maximum possible frictional force between the block and the incline?
If the block is released, what will be the net force in the x direction?
What therefore will be the acceleration of the block?
****
Estimated x component, 35%, estimated y component -90%.
Theta=307 degrees, mg=40 kg(9.8 m/s^2)=392 N
MgX=392Ncos(307)=332 N
MgY=392Nsin(307)=-313 N this is closer to 240 N
Fnormal+MgY=0, Fnormal+(-313N)=0, so Fnormal=313N
Ffriction=mu*Fnormal=.5*313N=157 N
FnetX=Ffriction+mgX=-157N+332N=175 N
Fnet=ma, 175 kgm/sec^2=40kg*a, a=4.4 m/sec^2
Good, but check out your calculation of the y component.
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`q010. In the preceding, what coefficient of friction would be required for friction alone to hold the block stationary?
****
I understand this concept, but Im not sure how to work backwards to find an exact value for mu in this situation
net force is zero implies friction equal and opposite to x comp of the weight
knowing frict force and normal force, you can calculate mu
See the given solution if necessary.
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`q011. Continuing the same situation, suppose the block is sliding up the incline, subject only to gravitational, normal and frictional forces. What would be the acceleration of the block?
****
Sliding up the incline means that Ffriction and MgX are working in the same direction, the negative direction. FnetX=Ffriction+mgX=-157N+-332N=-489N. Fnet=ma, -489 kgm/sec^2=40kg*a, a=-12.2 m/sec^2
very good; your frictional force is off a bit due to the miscalculation of Mg_y but that's only an arithmetic detail.
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ic_class_091026
Copy and paste this label into the form.
Report your work for today's class using the Submit Work Form. Answer the questions posed above.
** Note: Five very basic questions have been marked **. It is suggested you do these problems first, and you should make every effort to answer as many of those questions as possible prior to the next class. Most of the remaining questions are based on or at least related to these five. **
Read Chapter 4 of the text and take notes on things you do and do not understand. I might ask for a synopsis next time.
Work through and submit q_A_17 and q_A_18. These qa's are concerned primarily with vectors, and you should find them straightforward.
URL's of qa's 10-19:
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_11.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_12.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_13.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_14.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_15.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_16.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_17.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_18.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_19.htm
I've deleted the practice tests. If you did some work on them, you can resubmit. I didn't see any work, and didn't ask for any, but I don't want to miss anything you might have done.
Class 091021
short-answer questions
`q001. ** Give your answers to the following one-step questions:
What is the KE of a 10 kg object moving at 5 m/s?
What is `dPE for a 10 kg object raised 3 meters against the force of gravity?
What is `dPE for a block of dominoes pulled 15 cm in such a way to stretch a spring, if as a result the spring force changes linearly from 4 N to 16 N?
If on a certain interval, the KE and PE of a system change so that `dKE = + 8 J and `dPE = -12 J, then how much work is done by nonconservative forces on the system?
If on a certain interval nonconservative forces do 20 J of work on a system and its KE increases by 15 J, does the PE of the system increase or decrease, and by how much?
****
What is the KE of a 10 kg object moving at 5 m/s?
KE=.5mv^2=.5(10 kg)(5 m/s^2)=5 kg(25 m^2/s^2)=125 kgm^2/s^2=125 J
What is `dPE for a 10 kg object raised 3 meters against the force of gravity?
`dPE=-WorkONbyGravity=-Fgrav*`ds=-(m*a)*`ds=-(10 kg*9.8 m/s^2)*3 meters=-98 kgm/sec^2*3 m= 294 kgm^2/sec^2=294 J
What is `dPE for a block of dominoes pulled 15 cm in such a way to stretch a spring, if as a result the spring force changes linearly from 4 N to 16 N?
Since there is no change in vertical height, the effect of gravity does not change, and thus the potential energy of the domino block does not change.
If on a certain interval, the KE and PE of a system change so that `dKE = + 8 J and `dPE = -12 J, then how much work is done by nonconservative forces on the system?
`dWncON=`dPE+`dKE=-12 J+8 J=-4 J
If on a certain interval nonconservative forces do 20 J of work on a system and its KE increases by 15 J, does the PE of the system increase or decrease, and by how much?
`dWncON=`dPE+`dKE, 20 J=`dPE + 15 J, `dPE=5 J, so the potential energy increased by 5 Joules
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`q002. ** A slingshot shoots a pebble in the upward direction. While the slingshot is in contact with the pebble:
Does the elastic force exerted by the slingshot act in the direction of the pebble's motion or opposite this direction?
Does the elastic force therefore do positive or negative work on the pebble?
Does its elastic PE therefore increase or decrease?
Does the force of gravity act in the direction of the pebble's motion or opposite this direction?
Does the gravitational force therefore do positive or negative work on the pebble?
Does its gravitational PE therefore increase or decrease?
Assuming that the gravitational and elastic forces are the only conservative forces acting on the pebble, does its total PE increase or decrease?
****
The elastic force acts in the direction of the pebbles motion. I know this because the rubber of the slingshot travels forward with the pebble, but is stopped from completing its motion due to its attachment to the slingshot system.
The elastic force therefore does positive work on the pebble
The elastic PE increases. This can be proven as tension was put into the rubber band as the slingshot was pulled back before release.
The force of gravity works in the opposite direction of the pebbles outward motion, pulling it to the ground
The gravitational force therefore does negative work on the pebble
Gravitational PE decreases as the pebble starts from a high position and falls to the ground
The total potential energy of the pebble decreases completely. At first it had elastic potential energy behind it with the stretched rubber band as well as gravitational PE as it was higher than the ground. But after being projected forward, it lost contact with the rubber band and lost its elastic potential energy, and after being pulled all the way to the ground its gravitational potential energy was also reduced to zero.
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`q003. ** For the ball rolling in two directions along the 'constant-velocity' incline (i.e., an incline on which the velocity of the ball as it rolls down the incline remains unchanged), as you observed it in class:
What were your data?
Explain how you determined the acceleration of the ball the roll in each direction.
If the ramp was completely level, would it be a 'constant-velocity' incline?
If the ramp was completely level, what would be the acceleration in each direction?
On a 'constant-velocity' ramp, how are the parallel component of the gravitational force and the force of rolling friction related? (the 'parallel component' of the gravitational force is the component of that force in the direction of the incline)
****
The 61 cm ramp was nearly level, and a ball rolling from left to right had constant velocity.
A ball rolling from right to left, however, did not attain constant velocity.
It came to a stop at 44.5 cm in 10.4 sec, at 48.5 cm in 10.9 sec, at 53 cm in 11.6 seconds.
In the right direction, there was no acceleration as there was no change in velocity. A=`dv/`dt, and if `dv=0, a=0. In the left direction, acceleration can be found by using Vave=`ds/`dt, then v0=2*Vave (because vf=0), then a=`dv/`dt. I arrived at accelerations of .8 cm/sec^2, .82 cm/sec^2, and .79 cm/sec^2.
If the ramp was completely level, it would be considered a constant-velocity incline, but in reality the ball would need an initial push to get it started, creating an higher initial velocity, and it would slow to a stop by the end of the incline. The constant velocity would refer to the middle portion of the incline only.
If the ramp was completely level, the acceleration in each direction would be zero. This would be due to the zero change in velocity that I described above.
The positive work done by gravity in the X direction (level with the incline) would be equal and opposite to the negative work done by friction, slowing the ball down as it rolls.
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`q004. The rolling friction between a steel ball and a steel incline exerts a force opposed to the direction of motion. If this force is equal to .3% of the mass of the rolling ball, then
What is the frictional force on a 60 gram steel ball?
What would be the acceleration of the ball on a perfectly level ramp?
What would be its acceleration down a ramp with a 1 degree slope?
What would be its acceleration up the same ramp?
****
Ffriction=-.003(Fnormal)=-.003(.06 kg*9.8 m/sec^2)=-.003(.59 N)=-.002 Newtons
Acceleration on a perfectly level ramp would be zero
Let alpha=1 degree. MgX for this ramp would be .59 N*cos(1)=.59N. Fnet=Ffriction +MgX=-.002 N+.59N=.588 N. also, fnet=ma, so .588N=m*a=.06 kg*a, so a=9.8 m/sec^2
Up the same ramp, the ball would be traveling in the opposite direction, working in the same direction as the force of friction, so both forces would be considered negative. Fnet=Ffriction +MgX=-.002 N+-.59N=-.59 N. also, fnet=ma, so -.59N=m*a=.06 kg*a, so a=-9.9 m/sec^2
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`q005. The ball in the preceding rolls 50 cm down the 1 degree ramp, subject to only gravity, normal force and rolling friction. For this roll:
What is `dPE?
What is `dW_NC_ON?
What therefore is `dKE?
If the ball started from rest, what therefore is its final velocity?
****
`dPE=-WbyGravOn=-Fgrav*`ds=-(.59N)*.5 m=-.3N
I am unsure of how to solve for `dWncON without the `dWnet
If I had `dWncON, I would find `dKE either by using `dKE=-`dPE+`dWncON
I am unsure of how to solve for final velocity here.
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`q006. The acceleration of a 60-gram steel ball on a certain ramp with a small incline is observed to be 2.5 cm/s^2 when the ball moves in one direction, and -1.5 cm/s^2 when the ball moves in the opposite direction.
What is the net force on the ball as it moves in each direction?
Assuming that the force of rolling friction has the same magnitude in both directions, what is the force of rolling friction on the ball?
Is the ramp slopes in the positive or the negative direction?
What is the slope of the ramp?
****
One direction: fnet=ma=60 g (2.5 cm/sec^2)=150 gcm/sec^2, opposite direction: fnet=60 g(-1.5 cm/sec^2)=-90 gcm/sec^2
If rolling friction is the same in both directions, then the difference in forces in each direction are not due to friction, they are instead due to the ramp. This means ..?
Not sure how to calculate this
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`q007. ** My mass is about 75 kg. In about 40 minutes, my altitude and velocity change as follows:
I start at rest on a level runway in Atlanta, accelerate to 50 m/s before beginning to rise, rise to an altitude of 10 000 meters at a velocity of 200 m/s, descend to a level runway 300 miles away which I first contact at a speed of 60 m/s, and come to rest at the other end of the runway.
This scenario naturally breaks into five intervals, one occurring on the first runway, another between the end of that runway and cruising altitude, a third between cruising altitude and the beginning of descent, the fourth between the beginning of the descent and first contact with the second runway, and the last on the second runway.
Give quantitative answers to each of the following:
What happens to my gravitational potential energy during each phase?
What happens to my kinetic energy during each phase?
What nonconservative forces act during each phase?
What net work is done by nonconservative forces during each phase?
****
Gravitational PE does not change during phases 1, 3, or 5, at which the plane is moving parallel to the ground, not changing its altitude at all. In phase 2, gravitational PE increases as the plane works against gravity to obtain altitude. In Phase 4, gravitational PE decreases as the plane descends towards the ground, working with the force of gravity.
During Phase 1, kinetic energy increases as velocity increases. This can be explained by the direct relationship between KE and velocity in the equation KE= .5mv^2. This is also true of phase 2, at which velocity increases significantly. During phase 3 velocity is kept relatively constant, so there is no significant change in kinetic energy, but kinetic energy is still very high because the velocity of the plain is still very high. During phase 4 kinetic energy decreases as the plane slows, catching wind resistance as it falls through the air. During phase 5 kinetic energy returns to zero as the velocity of the plane comes back to zero.
In phase 1, air resistance and the turbines of the plane serve as nonconservative forces. The turbines are a positive force, while the air resistance is a negative force. The same is true of phases two and three and four. During phase five, air resistance plays a bigger part in slowing the plane, and the brakes of the plane work as a non conservative force as well, slowing the plane to a stop.
In phase one, the turbines do a significant amount of work while air resistance does little work against the plane as it is aerodynamic. The same is true of stage two and three. In phase 4 air resistance does much more work as the pilot positions the plane so it becomes less aerodynamic, therefore allowing it to slow and descend towards the runway. In phase 5 air resistance and the planes brakes do a significant amount of work in slowing the plane.
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`q008. A certain Atwood machine has 50 paperclips on each side. The paperclips have reasonably uniform mass. A positive direction has been declared, and the following observations are made.
If one of the clips is transferred to the positive side the system remains at rest.
If another clip is transferred to the positive side the system still remains at rest.
If a third clip is transferred to the positive side, the system accelerates at 15 cm/s^2.
After restoring the system to its original state, the system remains at rest after a transfer of 2 clips to the negative side.
A third clip transferred to the negative side results in an acceleration of - 10 cm/s^2.
First answer the following:
If there was no friction involved, what acceleration would result from transferring a single paperclip, so that one side has 51 clips and the other 49 clips?
****
It seems to me that it would matter which side of the system the paperclip was added to.
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Now, in terms of the stated data:
As a multiple of the weight of a paperclip, what do you conclude is the magnitude of the frictional force resisting the motion of the system?
****
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`q009. ** A 40 kg block rests on an incline which makes an angle of 37 degrees with the horizontal. The coefficient of friction between the block and the incline is mu = .5. Sketch a free-body diagram for this block, sketch x and y axes with the x axis parallel to the ramp, and estimate the x and y components of the weight of the block. Then using the sine and cosine functions, calculate these components.
Give your estimates of the x and y components, as well as the calculated x and y components of the weight.
Using the fact that the block remains in equilibrium in the y direction, determine the normal force between block and incline.
What therefore is the maximum possible frictional force between the block and the incline?
If the block is released, what will be the net force in the x direction?
What therefore will be the acceleration of the block?
****
Estimated x component, 35%, estimated y component -90%.
Theta=307 degrees, mg=40 kg(9.8 m/s^2)=392 N
MgX=392Ncos(307)=332 N
MgY=392Nsin(307)=-313 N
Fnormal+MgY=0, Fnormal+(-313N)=0, so Fnormal=313N
Ffriction=mu*Fnormal=.5*313N=157 N
FnetX=Ffriction+mgX=-157N+332N=175 N
Fnet=ma, 175 kgm/sec^2=40kg*a, a=4.4 m/sec^2
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`q010. In the preceding, what coefficient of friction would be required for friction alone to hold the block stationary?
****
I understand this concept, but Im not sure how to work backwards to find an exact value for mu in this situation
&&&&
`q011. Continuing the same situation, suppose the block is sliding up the incline, subject only to gravitational, normal and frictional forces. What would be the acceleration of the block?
****
Sliding up the incline means that Ffriction and MgX are working in the same direction, the negative direction. FnetX=Ffriction+mgX=-157N+-332N=-489N. Fnet=ma, -489 kgm/sec^2=40kg*a, a=-12.2 m/sec^2
&&&&
ic_class_091026
Copy and paste this label into the form.
Report your work for today's class using the Submit Work Form. Answer the questions posed above.
** Note: Five very basic questions have been marked **. It is suggested you do these problems first, and you should make every effort to answer as many of those questions as possible prior to the next class. Most of the remaining questions are based on or at least related to these five. **
Read Chapter 4 of the text and take notes on things you do and do not understand. I might ask for a synopsis next time.
Work through and submit q_A_17 and q_A_18. These qa's are concerned primarily with vectors, and you should find them straightforward.
URL's of qa's 10-19:
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_11.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_12.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_13.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_14.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_15.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_16.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_17.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_18.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_19.htm
In preparation for the upcoming test, you should be sure you understand the problems and solutions in the Introductory Problem Sets 3, 4 and 5. For Phy 121 students, the test is generated almost entirely from these problems. For Phy 201 students, about 50% of the test comes from these problems.
The following will be used as sample tests (tests are included for both Phy 121 and Phy 201). It is recommended that you at least read over these tests before the next class. It is not expected that everyone will have time to work through the tests before the next class.
It appears that you submitted two copies of the assignment. My comments are in the first; if the second differs from the first let me know. Otherwise no harm done.
Below is a copy of the 'given solutions'.
class 091026
Class 091021
short-answer questions
`q001. ** Give your answers to the following
one-step questions:
- What is the KE of a 10 kg object moving at 5 m/s?
- What is `dPE for a 10 kg object raised 3 meters
against the force of gravity?
- What is `dPE for a block of dominoes pulled 15 cm in
such a way to stretch a spring, if as a result the spring force changes
linearly from 4 N to 16 N?
- If on a certain interval, the KE and PE of a system
change so that `dKE = + 8 J and `dPE = -12 J, then how much work is done by
nonconservative forces on the system?
- If on a certain interval nonconservative forces do 20
J of work on a system and its KE increases by 15 J, does the PE of the
system increase or decrease, and by how much?
****
The KE of a 10 kg object moving at 5 m/s is 1/2 m v^2 = 1/2 *
10 kg * (5 m/s)^2 = 5 kg * 25 m^2 / s^2 = 125 kg m^2/s^2 = 125 J.
`dPE of a system is equal and opposite to work done by a
conservative force acting on the system. The conservative force here is
gravity, which exerts a downward force of 10 kg * 9.8 m/s^2 = 98 N as the object
moves 3 meters in the upward direction. Thus gravity does negative work:
`dW_grav_ON = -98 N * 3 m = -294 J
and the PE change is equal and opposite
`dPE = -`dW_grav_ON = - (-294 J) = 294 J.
The force increases linearly, so the average force is the
average of the initial and final forces, in this case (4 N + 16 N) / 2 = 10 N.
The displacement is 15 cm opposite the direction of the force (the object moves
one way, the conservative acting ON the object force pulls back the other way),
so conservative force does work -10 N * .15 m = -1.5 N * m = -1.5 J.
If `dKE = +8 J and `dPE = -12 J, then
dW_NC_ON = `dKE+ `dPE = +8 J + (-12
J) = -4 J.
`dW_NC_ON = `dKE + `dPE, so `dPE = dW_NC_ON - `dKE = 20 J - 15
J = 5 J.
&&&&
`q002. ** A slingshot shoots a pebble in the upward
direction. While the slingshot is in contact with the pebble:
- Does the elastic force exerted by the slingshot act
in the direction of the pebble's motion or opposite this direction?
- Does the elastic force therefore do positive or
negative work on the pebble?
- Does its elastic PE therefore increase or decrease?
- Does the force of gravity act in the direction of the
pebble's motion or opposite this direction?
- Does the gravitational force therefore do positive or
negative work on the pebble?
- Does its gravitational PE therefore increase or
decrease?
- Assuming that the gravitational and elastic forces
are the only conservative forces acting on the pebble, does its total PE
increase or decrease?
****
The elastic force exerted by the slingshot as it 'snaps back'
toward equilibrium is in the upward direction, as is the displacement of the
pebble.
The elastic force therefore does positive work on the pebble.
It follows that the change in elastic PE of the system is
negative, so the elastic PE decreases.
The gravitational force acts downward while the pebble rises.
The gravitational force therefore does negative work on the
pebble.
Its gravitational PE therefore decreases.
The slingshot manages to 'shoot' the pebble upwards,
increasing its KE. Whether we assume that `dW_NC_ON is zero or negative,
we conclude that `dPE = `dW_NC_ON - `dKE must be negative.
&&&&
`q003. ** For the ball rolling in two directions
along the 'constant-velocity' incline (i.e., an incline on which the velocity of
the ball as it rolls down the incline remains unchanged), as you observed it in
class:
- What were your data?
- Explain how you determined the acceleration of the
ball the roll in each direction.
- If the ramp was completely level, would it be a
'constant-velocity' incline?
- If the ramp was completely level, what would be the
acceleration in each direction?
- On a 'constant-velocity' ramp, how are the parallel
component of the gravitational force and the force of rolling friction
related? (the 'parallel component' of the gravitational force is the
component of that force in the direction of the incline)
****
Suppose that the ball traveled 40 cm 'up' the incline,
requiring 8.5 half-swings of a 16-cm pendulum to come to rest (traveling
'down' the incline, presumably, it has zero acceleration and if starts does
not come to rest). Then the period of the pendulum is .2 sqrt(16) s =
.8 s, and 8.5 half-swings require 6.8 sec. The average velocity on the
incline is
v_Ave = `ds / `dt = 40 cm / (6.8 s) = 6 cm/s, approx..
Since final velocity is zero, assuming acceleration to be
constant we see that the initial velocity must be double the average
velocity, so
v0 = 2 * 6 cm/s = 12 cm/s.
The change in velocity is therefore `dv = -12 cm/s, and
the acceleration is
a = `dv / `dt = 12 cm/s / (6.8 s) = 1.7 cm/s^2, approx..
The difference between the 0 acceleration 'down' the ramp
and the 1.7 cm/s^2 acceleration 'up' the ramp is due to friction. The
ramp has sufficient slope that the 'parallel component' of the gravitational
force exactly counters the frictional force. When the ball travels
'up' the ramp it encounters the 'parallel component' of the gravitational
force, which now resists its motion, as well as the frictional force.
These forces are equal, so each alone would produce an acceleration of
magnitude 1/2 * 1.7 cm/s^2 = .85 cm/s^2.
If the ball was rolled along a perfectly level ramp, the
'parallel component' of its acceleration would be zero, and the net force
would be just the frictional force. Its acceleration would therefore
be .85 cm/s^2, in the direction opposite its motion. Rolled in
opposite directions, then, its accelerations might therefore be -.85 cm/s^2
and +.85 cm/s^2. The difference in the accelerations would still be
1.7 cm/s^2, as was the case for the constant-velocity ramp.
&&&&
`q004. The rolling friction between a steel ball and
a steel incline exerts a force opposed to the direction of motion. If this
force is equal to .3% of the weight of the rolling ball, then
- What is the frictional force on a 60 gram steel ball?
- What would be the acceleration of the ball on a
perfectly level ramp?
- What would be its acceleration down a ramp with a 1
degree slope?
- What would be its acceleration up the same ramp?
****
The weight of the ball is the force exerted on it by gravity,
which is the product of its mass and the acceleration of gravity. The
magnitude of the weight is therefore
weight = m g = .060 kg * 9.8 m/s^2 = .6 N, approx.
The frictional force is .3% of this, or
f_frict = .003 * .6 N = .0018 N.
On a perfectly level ramp, the net force will be the
frictional force, so the acceleration of the ball will be
a = F_net / m = .0018 N / (.060 kg) = .03 m/s^2.
On a 1% incline, the gravitational force will make an angle
of 271 deg, provided the positive x axis is oriented down the ramp.
The x component of the gravitational force (which is parallel to the
incline, hence the name 'parallel component') is therefore
x comp of wt = wt * cos(271 deg) = .6 N * .017 = .011 N,
approx.
and the net force on the ball is
F_net = wt_x + f_frict = .011 N + (-.0018 N) = .009 N, where
the positive direction is taken to be down the incline (the ball accelerates
in the positive direction, friction being opposite the motion down the ramp
is negative)
Its acceleration will therefore be
a = F_net / m = .009 N / (.060 kg) = .15 m/s^2, down the
incline.
Moving up the incline the parallel component of the
gravitational force and the frictional force are both down the incline, so
keeping the earlier choice of positive direction the net force is
F_net = .011 N + .0018 N = .013 N, approx.
and the acceleration is
a = F_net / m = .013 N / (.060 kg) = ,22 m/s^2, approx..
So a ball travcling down the ramp (in the direction we have
chosen as positive)( experiences net force .009 N down the ramp, and
accelerates at .15 m/s^2 down the incline, which speeds it up. A ball
traveling up the ramp (now in our chosen negative direction) experiences a
net force of .013 N and an acceleration of .22 m/s^2, which slows it down.
&&&&
`q005. The ball in the preceding rolls 50 cm down
the 1 degree ramp, subject to only gravity, normal force and rolling friction.
For this roll:
- What is `dPE?
- What is `dW_NC_ON?
- What therefore is `dKE?
- If the ball started from rest, what therefore is its
final velocity?
****
Rolling 50 cm down the ramp with the
'parallel' gravitational force wt_x = .011 N, the gravitational force does
work
`dW_grav_ON = wt_x * ds_x =
,011 N * .050 m = .00055 J
on the ball, resulting in
`dPE = - `dW_grav_ON = -.00055 J.
The nonconservative force is the
frictional force, which is .0018 N in the direction opposite motion, so it
does negative work:
`dW_NC_ON = -.0018 N * .050 m = -.00009
J.
Its KE change is therefore
`dKE = `dW_NC_ON - `dPE = -.00009 J -
(-.00055 J) = .00046 J
If the ball starts from rest, then its
initial KE is 0 so its final KE is
KE_f = KE_0 + `dKE = 0 + .00046 J =
.00046 J.
&&&&
`q006. The acceleration of a 60-gram steel ball on a
certain ramp with a small incline is observed to be 2.5 cm/s^2 when the ball
moves in one direction, and -1.5 cm/s^2 when the ball moves in the opposite
direction.
- What is the net force on the ball as it moves in each
direction?
- Assuming that the force of rolling friction has the
same magnitude in both directions, what is the force of rolling friction on
the ball?
- Is the ramp slopes in the positive or the negative
direction?
- What is the slope of the ramp?
****
Moving in the positive direction the acceleration is 2.5
cm/s^2, so
F_net = .060 kg * .025 m/s^2 = .0015 N, approx.
Moving in the negative direction the acceleration is -1.5
cm/s^2, so
F_net = .060 kg * (-.015 m/s^2) = -.0009 N, approx..
The net force differs because friction acts in opposite
directions when the ball moves in opposite directions. All other
forces are the same. So the difference in the two forces is double the
force of friction.
The difference between the two forces is .0015 N - (.0009
N) = .0024 N, so the frictional force is half this:
f_frict = difference between forces / 2 = .0024 N / 2 =
.0012 N.
The average of the two forces is (.0015 N + (-.0009 N) /
2 = .0003 N.
This is the 'parallel component' of the weight, i.e.
wt_x = .0003 N.
The weight of the ball is .06 N.
You can use the Pythagorean Theorem to find the y
component of the weight; since the x component is much less than the weight,
the y component will differ very slightly from the .06 N weight; the y
component of the weight will of course be negative.
So the weight vector has x component .0003 N and y
component .06 N. Its angle with the positive x axis is therefore
angle = arcTan ( .06 N / .0003 N) = arcTan(200) = -89.7
degrees, approx..
Thus the weight vector lies .3 degrees from the negative
y axis, so the incline lies .3 degrees above horizontal.
&&&&
`q007. ** My mass is about 75 kg. In about 40
minutes, my altitude and velocity change as follows:
- I start at rest on a level runway in Atlanta,
accelerate to 50 m/s before beginning to rise, rise to an altitude of 10 000
meters at a velocity of 200 m/s, descend to a level runway 300 miles away
which I first contact at a speed of 60 m/s, and come to rest at the other
end of the runway.
This scenario naturally breaks into five intervals, one
occurring on the first runway, another between the end of that runway and
cruising altitude, a third between cruising altitude and the beginning of
descent, the fourth between the beginning of the descent and first contact with
the second runway, and the last on the second runway.
Give quantitative answers to each of the following:
- What happens to my gravitational potential energy
during each phase?
- What happens to my kinetic energy during each phase?
- What nonconservative forces act during each phase?
- What total work is done by nonconservative forces
during each phase?
****
During the first phase, altitude does not change so
gravitational PE does not change; i.e., `dPE = 0. KE changes from 0 to
1/2 * 75 kg * (50 m/s)^2 = 9400 J, approx. The nonconservative forces
acting are air resistance, which acts in the direction opposite motion and
therefore does negative work, rolling friction, which similarly does
negative work, and the force exerted on the airplane by the air expelled
from the jet engines. The latter is equal and opposite to the force
exerted by the engines on the air. The air is expelled toward the back
of the aircraft, which requires a backward force. The force of the
expelled air on the engines is therefore forward, in the direction of
motion, and does positive work on the airplane.
During this phase `dW_NC_ON = `dKE + `dPE = 9400 J + 0
J = 9400 J. This is the sum of the positive work done in reaction to
the force exerted by the engines, and the negative work done by air
resistance. In other words, the engines have to supply the 9400 J to
accelerate my mass, in addition to my share of the air resistance.
During the second phase, KE changes from about 9400 J
to 1/2 * 75 kg * (200 m/s)^2 = 1 500 000 J and gravitational PE changes by
75 kg * 9.8 m/s^2 * 10 000 m = 7 500 000 J, approx.. The
nonconservative forces are air resistance and the force exerted by the
expelled air on the engines.
During this phase `dW_NC_ON = `dKE + `dPE = 1 500 000
J + 7 500 000 J = 9 000 000 J. This energy, plus the energy required
to overcome air resistance, are supplied by the fuel that runs the engines.
During the third phase neither the speed of the
airplane nor its altitude change, so neither KE nor PE change.
During this phase `dW_NC_ON = `dPE + `dKE = 0.
The negative work done by the air resistance and the positive work done by
the air expelled by the engines are equal and opposite.
During the fourth phase the PE changes by -7 500 000
J, approx., and KE decreases to 1/2 * 75 kg * (60 m/s)^2 = 13 500 J.
During this phase `dW_NC_ON = `dPE + `dKE = -7 500 000
J - 1 500 000 J = -9 000 000 J. This is the sum of the negative work
done by air resistance and the positive work done by the engines. To
get to the runway with a survivable speed, the airplane must position itself
in such a way as to intercept a lot of air. Remember, that -9 000 000
J is just for my mass.
Between contacting the runway and coming to rest, PE
does not change and KE decreases from 13 500 J to 0.
Thus `dW_NC_ON = `dPE + `dKE = 0 - 13 500 J = -13 500
J. Air resistance won't accomplish this; the airplane relies on
brakes, friction and reversing its engines to come to rest.
&&&&
`q008. A certain Atwood machine has 50 paperclips on
each side. The paperclips have reasonably uniform mass. A positive
direction has been declared, and the following observations are made.
If one of the clips is transferred to the positive side
the system remains at rest.
If another clip is transferred to the positive side the
system still remains at rest.
If a third clip is transferred to the positive side, the
system accelerates at 15 cm/s^2.
After restoring the system to its original state, the
system remains at rest after a transfer of 2 clips to the negative side.
A third clip transferred to the negative side results in
an acceleration of - 10 cm/s^2.
First answer the following:
- If there was no friction involved, what acceleration
would result from transferring a single paperclip, so that one side has 51
clips and the other 49 clips?
****
The system has mass 100 * m_clip, where m_clip is the
mass of a single clip.
With 50 clips on each side, the positive and negative
forces are equal, and the net force is zero. The system does not
accelerate.
With 51 clips on one side and 49 clips on the other,
the downward forces on the two masses are 51 * m_clip * g and 49 * m_clip *
g. One force pulls the system in the positive direction and the other
in the negative direction, so the net force has magnitude
| F_net | = 51 m_clip * g - 49 * m_clip * g = 2 m_clip
* g.
That is, the net force is equal to the force exerted
by gravity on 2 clips.
The magnitude of the acceleration is
| a | = | F_net | / m_system = 2 m_clip * g / (100
m_clip * g) = .02 * g.
That is, the force exerted by gravity on 2 clips will
accelerate a system with a mass equal to that of 100 clips at 2/100 the
acceleration of gravity.
.02 * g = .196 m/s^2 (easily calculated from g = 9.8
m/s^2).
&&&&
Now, in terms of the stated data:
- As a multiple of the weight of a paperclip, what do
you conclude is the magnitude of the frictional force resisting the motion
of the system?
****
&&&&
`q009. ** A 40 kg block rests on an incline which
makes an angle of 37 degrees with the horizontal. The coefficient of
friction between the block and the incline is mu = .5. Sketch a
free-body diagram for this block, sketch x and y axes with the x axis parallel
to the ramp, and estimate the x and y components of the weight of the block.
Then using the sine and cosine functions, calculate these components.
- Give your estimates of the x and y components, as
well as the calculated x and y components of the weight.
- Using the fact that the block remains in equilibrium
in the y direction, determine the normal force between block and incline.
- What therefore is the maximum possible frictional
force between the block and the incline?
- If the block is released, what will be the net force
in the x direction?
- What therefore will be the acceleration of the block?
****
Your sketch should depict the x-y axes rotated 37
degrees, either clockwise or counterclockwise, from the 'standard'
vertical-horizontal orientation. The vertical weight vector will
therefore lie 37 degrees from the negative x axis.
Assuming counterclockwise rotation, with the ramp
inclined up and to the right, the weight vector will lie in the third
quadrant, 37 degrees from the negative y axis, at angle 270 deg - 37 deg =
233 deg counterclockwise from the positive x axis. (If your ramp is
inclined down and to the right the rotation of the axes will be clockwise
and the angle will be 270 deg + 37 deg = 307 deg.)
The weight is 40 kg * 9.8 m/s^2 = 400 N, approx..
The components are therefore
wt_x = 400 N cos(233 deg) = -240 N and
wt_y = 400 N sin(233 deg) = -320 N
(if your ramp is inclined down and to the right your
angle is 307 deg and the x component will be positive; the y component will
be the same).
The only forces acting in the y direction are the y
component of the weight and the normal force. Since we have y
equilibrium (no acceleration in the y direction so zero net force in the y
direction), these forces must be equal and opposite:
F_normal + wt_y = 0 so
F_normal = -wt_y = - ( -320 N) = 320 N.
The coefficient of friction is mu = .5. So the
maximum possible frictional force has magnitude
| f_frict_max | = .5 * F_normal = .5 * 320 N = 160 N.
The x component of the weight, which acts parallel to the
incline, has magnitude 240 N. Frictional force, which cannot exert
more that 160 N, acts in the direction this force. So the net force is
in the x direction, down the incline, and has magnitude
| F_net | = 240 N - 160 N = 80 N.
The net force is therefore 80 N down the incline.
The acceleration of the block is therefore
a = F_net / m = 80 N / (40 kg) = 2 m/s^2, down the
incline.
&&&&
`q010. In the preceding, what coefficient of
friction would be required for friction alone to hold the block stationary?
****
To hold the block stationary would require a frictional
force of 240 N, equal and opposite to the 'parallel component' of the weight.
The normal force is 320 N, and the frictional force cannot
be greater than mu * F_normal. Therefore
mu * F_normal >= 240 N and
mu >= 240 N / (F_normal) = 240 N / (320 N) = .75.
The coefficient of friction must be at least .75.
&&&&
`q011. Continuing the same situation, suppose the
block is sliding up the incline, subject only to gravitational, normal and
frictional forces. What would be the acceleration of the block?
****
If the block is sliding up the incline then the frictional
force (160 N) is acting down the incline, in the same direction as the 'parallel
component' of the weight (240 N).
So the magnitude of the net force will be
| F_net | = 160 N + 240 N = 400 N.
The acceleration of the system therefore has magnitude
| a | = | F_net | = 400 N / (40 kg) = 10 m/s^2.
&&&&