#$&*
course Phy 201
12:20 am sept 15Sorry for the tardiness of this assignment. Life has been crazy. I'll work hard to make sure to turn in my remaining assignments on time.
I have a few questions on here where I didn't even know where to start so hopefully you can help me out :)
Thanks so much! " "Applying basic definitions of motion, force, work and energy to some of our fundamental systems
--------------------------------------------------------------------------------
A lot of this document can be regarded as sort of a reading comprehension exercise.
If you interpret the words right you can get the right answers, even if you're not completely sure what some of the answers mean.
The main task is to figure out what to do with the given information, in terms of the relationships you have been given, and do the appropriate calculations (being
sure to include the units; however you aren't asked to interpret the units and you may simply express the units as they are given).
Another important task is to get used to the ideas of motion, force, work-energy, etc. in the context of some of the systems we have observed.
--------------------------------------------------------------------------------
Information from preceding assignments:
Rate of change:
The average rate of change of A with respect to B is (change in A) / (change in B).
Average velocity is average rate of change of change of position with respect to clock time.
Average acceleration is average rate of change of position with respect to clock time.
Graph Trapezoids
The slope of a 'graph trapezoid' is its rise / run.
The 'graph altitudes' of a 'graph trapezoid' are the quantities represented by its vertical sides.
The area of a 'graph trapezoid' is its average 'graph altitude' multiplied by its width.
Force, work-energy, momentum
When an object of mass m is moving with velocity v, it has the following properties
its kinetic energy is KE = 1/2 m v^2
its momentum is p = m v
Forces acting on objects can change their velocity, momentum and kinetic energy.
When an object of mass m changes its velocity, with respect to clock time, at rate a, then the net force acting on it (i.e., the sum of all the forces acting on it) is
F_net = m * a.
If a force F acts through a displacement `ds along the line of the force, then the force does work `dW = F * `ds.
If F in the above happens to be the net force acting on an object, then the KE of that object changes by an amount equal to `dW.
If a net force F_net acts on an object for time interval `dt, then the momentum of that object changes by `dp = F_net * `dt.
--------------------------------------------------------------------------------
As on the preceding assignment, some questions can be answered fairly directly, while others are more challenging. Don't let yourself get bogged down on any one
question before moving on to another. Come back on another day to questions you can't answer on your first try.
`q001. On a graph of velocity v (in cm/sec) vs. clock time t (in sec):
What are the velocity and clock time corresponding to the point (4, 12)?
****
V = 12 cm/sec
clock time = 4 sec
#$&*
What are the velocity and clock time corresponding to the point (9, 32)?
****
V = 32 cm/sec
clock time = 4 sec
#$&*
If these points correspond to the velocity of a ball rolling down an incline, describe as fully as you can what you think happens between the first event
(corresponding to the first point of the graph) and the second event (corresponding to the second point of the graph).
****
the acceleration of the ball increases between the first event and the second
a is represented with a positive slope indicating that it is increasing
#$&*
What is the change in velocity between these two events?
****
delta V = (32 cm/sec) - (12 cm/sec) = 20 cm/sec
#$&*
What is the change in clock time between these two events?
****
delta clock time = 9 sec - 4 sec = 5 sec
#$&*
What is the average velocity for the interval between these two events?
****
(32 cm/sec + 12 cm/sec) / 2 = (44 cm/sec) / 2 = 22 cm/sec
#$&*
What is the average rate at which the velocity changes, with respect to clock time, between these two events?
****
Quantity A = velocity
Quantity B = Clock time
rate of change = (20 cm/sec) / (5 sec) = 4 cm/sec^2
#$&*
What is the displacement of the object between these two events?
****
travels 5 seconds between event 1 and event 2
average velocity = 22 cm/sec
22cm / 1sec = 110cm/5sec
displacement = 110 cm in 5 sec
#$&*
`q002. A 5 kg mass accelerates at 2 m/s^2. What is the net force acting on the object?
****
F_net = m * a = 5kg * 2m/sec^2 = 10 kg*m/sec^2
#$&*
`q003. A net force of 5000 kg m/s^2 acts on a 100 kg mass. What is the acceleration of the mass?
****
F_net = m * a
500 kg = 5kg * a
a = 1000 m/s^2
#$&*
`q004. A force of 400 Newtons is exerted on an automobile as it is pushed through a distance of 100 meters. How much work is done on the automobile?
****
'dW = F * 'd = 400 N * 100 m = 40000 N*m
#$&*
`q005. A certain pendulum requires .5 Newtons of force for every centimeter it is pulled back (recall pulling back the pendulum hanging from the tree limb, using the
rubber band).
How much force would be required to pull the pendulum back 5 cm, 10 cm and 15 cm?
****
.5 N required for every cm
5 cm * .5N = 2.5 N cm
10 cm * .5N = 5 N cm
15 cm * .5 N = 7.5 N cm
#$&*
What is the average force required between pullbacks of 5 cm and 15 cm?
****
F_ave = (2.5 N cm + 7.5 N cm) / 2 = 5 N cm
#$&*
How much work is done between these two positions?
****
'dW = F * 'ds = 5 N cm * 10 cm = 50 N cm^2
#$&*
`q006. A system rotates through 12 rotations in 4 seconds, first coming to rest at the end of this interval.
How quickly is it rotating, on the average? (The answer is as obvious as it should seem, but also be sure to interpret this as a rate of change with respect to clock
time, and carefully apply the definition of average rate)
****
rotaions A
clock time B
12 rotations / 4 sec = 3 rotations / sec
#$&*
Is it speeding up or slowing down?
****
assuming that it started from rest and came to rest, then it was initially speeding up and then started to slow back down
#$&*
At what rate is it doing so? (Again attempt to interpret as an average rate of change of an appropriate quantity with respect to clock time).
****
average rate of change in velocity (A) wrt clock time (B)
(3 rotations / sec) / 4 sec = (3 rotations / sec) * (1 / 4 sec) = 3 rotations / 4 sec^2
#$&*
@& Very good.
The change is actually -3 rot / sec, and if you do the arithmetic you get
-3 rot / (4 sec^2) = -.75 rot / sec^2.
Your solution is the closest I've seen to the correct answer to this question, which seems to be eluding most students. That's no surprise at this point, but I'm glad to see that you're getting it.*@
`q007. At the initial point of an interval a 7 kg mass is moving at 5 meters / second. By the end of the interval it has gained an additional 200 kg m^2 / s^2 of
kinetic energy.
How much kinetic energy does it therefore have at the end of the interval?
****
KE = 1/2 m v^2 = 1/2 (7kg) (5m/sec)^2 = 3.5kg * 25m^2/s^2 = 87.5kg*m^2/sec^2 = 200 kg*m^2 = 287.5 kg*m^2/s^2
#$&*
Slightly more challenging question: How fast is the mass therefore moving at the end of the interval?
****
287.5 kg*m^2/s^2 = 1/2 (7kg) (v^2)
multiply both sides by 2
575 kg*m^2/sec^2 = 7kg * v^2
divide by 7 kg
82.1425714 m^2/s^2 = v^2
take the square root of both sides
V ~= 9.1 m/s
#$&*
`q008. An object begins an interval with a kinetic energy of 20 000 kg m^2 / s^2, and ends the interval with a kinetic energy of 15 000 kg m^2 / s^2.
By how much did its kinetic energy change on this interval? (The answer is as obvious as it might seem, but be careful about whether the answer is positive or
negative).
****
between beginning and end, it changed -5000kg*m^2/s^2
#$&*
More challenging: During this interval, how much work was done on the object by the net force?
****
I know the KE, from that I could get m & v, if I could get V then I could find a, then I could find the F_net by multiplying m & a ttogether, with the F_net, I would
multiply it by the 'ds to find the 'dW
#$&*
@& Excellent thinking, but in this case you need only recall that the change in KE is equal to the work done by the net force.*@
Also more challenging: If the average force on the object during this interval had magnitude 200 Newtons, then what was its displacement during this interval?
****
F_net = 200 N
'ds = 'dW / F = 'dW / 200 N
F_net = m * a
'dW / 'ds = m * a = 200 N
#$&*
@& Right. You do have the means to get the work (see my preceding note).*@
`q009. A mass of 200 grams hangs from one side of a pulley, and another mass from the other side. The gravitational force pulling down on this mass is about 200 000
gram cm / s^2, and the tension in the string pulling it upward is about 180 000 gram cm / s^2.
Pick either upward or downward as the positive direction.
****
up
#$&*
Using + for your positive direction and - for your negative direction, what is the gravitational force on this object?
****
-
#$&*
@& This question asked for the grav force, not just the sign of that force.*@
Using + for your positive direction and - for your negative direction, what is the tension force on this object?
****
+
#$&*
Using + for your positive direction and - for your negative direction, what is the net force on this object?
****
-
#$&*
Using + for your positive direction and - for your negative direction, what is the acceleration of this object?
****
+
#$&*
If the object's displacement during a certain interval is +30 cm, then according to your choice of positive direction, is the displacement upward or downward?
****
up
#$&*
When you multiply the displacement by the gravitational force, what is your result? Be sure to indicate whether the result is positive or negative.
****
'ds * F_grav
+30 * -200000 gsm/s^2 = -6000000gcm^2/s^2
#$&*
When you multiply the displacement by the tension force, what is your result? Be sure to indicate whether the result is positive or negative.
****
+30 * 180000gcm/S^2 = 5400000gcm^2/s^2
#$&*
When you multiply the displacement by the net force, what is your result? Be sure to indicate whether the result is positive or negative.
****
+30 cm * -200000gcm/sec^2 = -19970gcm^2/s^2
#$&*
@& The net force is the total of the gravitational and tension forces, and is just -20 000 g cm/s^2.*@
Does gravity do positive or negative work on this object?
****
-
#$&*
Does the tension force do positive or negative work on this object?
****
+
#$&*
Does the net force do positive or negative work on this object?
****
-
#$&*
Does the object speed up or slow down?
****
from rest: speeds up
to rest: slows down
#$&*
How would your answers have changed if you had chosen the opposite direction as positive?
****
it would have made more sense with the numbers that were provided
the system would still be moving in the say way
#$&*
@& The problem makes as much sense with the positive direction up as with the positive direction down. Really no difference; one choice is as good as the other.
The displacements and forces would all have had the opposite sign.
However the work by each force would have been the same, as would the description of motion in terms of up and down.*@
`q010. A pendulum of length 2 meters and mass 3 kg, pulled back a distance | x | from its equilibrium position, experiences a restoring force of magnitude k | x |,
where k = 15 kg / s^2 * | x |. [Note that for convenience in calculation we are making some approximations here; the actual value of k for this pendulum would
actually be closer to 14.7 kg / s^2, and this is so only for values of | x | which are a good bit smaller than the length. These are details we'll worry about later.]
How much force does the pendulum experience when x = .1 meter?
****
15kg/s^2 * .1m = 1.% kg m/s^2
#$&*
How much force does the pendulum experience when x is .05 meter, .1 meter, .15 meter and .2 meter?
****
15kg/s^2 * .05 = 0.75 kg m / s^2
15kg/s^2 * .1 = 1.5 kg m / s^2
15kg/s^2 * .15 = 2.25 kg m / s^2
15kg/s^2 * .2 = 3 kg m / s^2
#$&*
What do you think is the average force between | x | = .05 meter and x = .2 meter?
****
(0.75 kg m / s^2 + 3 kg m / s^2) / 2 = 1.875 kg m / s^2
#$&*
How much work do you think would be done by this force between | x | = .05 meter and | x | = .2 meter?
****
1.875 kg m / s^2 * .15 m = 0.28125 kg m^2 / s^2
#$&*
How fast would the pendulum have to be going in order for its kinetic energy to equal the result you just obtained for the work?
****
KE + o.28125 kg m^2 / s^2 = 1/2 m * v^2
V = 0.4330127019
#$&*
If the pendulum moves from position x = .05 meter to x = .2 meter, is the direction of the force the same as, or opposite to the direction of the motion?
****
same?
~Need help with this concept~
I drew a picture, but i ended up with the same thing for this and the next so...
#$&*
@& If you pull a pendulum back does the pendulum feel like it's pulling back toward the equilibrium position or away from it?
Is the motion from one point to the other, in this question, toward or away from equilibrium?*@
If the pendulum moves from position x = .20 meter to x = ..05 meter, is the direction of the force the same as, or opposite to the direction of the motion?
****
same?
#$&*
If the pendulum string was cut, what would be the acceleration of the 1 kg mass?
****
I know that the acceleration of an object is the change in its velocity divided by the change in time
regardless of this knowledge, I'm afraid I don't understand this question
#$&*
@& If the string is cut the string stops exerting a force and the pendulum is subject only to the force of gravity, which accelerates it downward at 9.8 m/s^2.*@
What is the magnitude of the force exerted by gravity on the pendulum's mass? For simplicity of calculation you may use 10 m/s^2 for the acceleration of gravity.
****
F_net = m * a = 1kg * 10 m/s^2 = 10 kg m / s^2
#$&*
@& Good, but the mass of the pendulum was originally given as 3 kg.*@
When x = .1 meter, what is the horizontal displacement from equilibrium as a percent of the pendulum's length?
****
.1 m / 2 m * 100 = 5%
#$&*
When x = .1 meter, what is the restoring force as a percent of the pendulum's weight?
****
Is the restoring force the force that is needed to bring the pendulum back to rest
#$&*
@& Yes. That's the 1.5 Newtons you found earlier for this position.*@
What is the magnitude of the acceleration of the pendulum at the x = .15 meter point?
****
I know the equations and the concepts but this one is still stumping me
#$&*
@& You were given the mass of the pendulum, and you found the force at the .15 meter point. So you can find the acceleration.*@
`q011. The force exerted on a mass has magnitude | F | = 15 Newton / meter * x, for 0 <= x <= .25 meter.
Sketch a graph of | F | vs. x. (You might wish to start by making a table of | F | vs. x, for some appropriate values of x between 0 and .25 meter). Note the
convention that a graph of y vs. x has y on the vertical axis and x on the horizontal, so that for this graph | F | will be on the vertical axis and x on the
horizontal.
****
okay
#$&*
Verify that the points (.06 meter, .9 Newton) and (.16 meter, 2.4 Newtons) lie on your graph.
****
15 N / 1 m = .9 N / .06 m = 2.4 N / .16 m
#$&*
What is the rise between these points?
****
1.5 N
#$&*
What is the run between these points?
****
.1 m
#$&*
What is the average slope between these points?
****
1.5 N / .1 m = 15 N/m
#$&*
What is the average 'graph altitude' of the 'graph trapezoid' formed by these points?
****
(.9 + 2.4) / 2 = 1.65 N
#$&*
What is the width of the trapezoid?
****
0.16 - 0.06 = .1 m
#$&*
What therefore is the area of the trapezoid?
****
1.65 N * .1 m = .165 N*m
#$&*
What are the graph points corresponding to x = .05 meter and to x = .20 meter?
****
(0.05 , 0.75)
(.2 , 3)
#$&*
What is the area of the 'graph trapezoid' defined by these points?
****
1.875 N * 0.03 m = 0.05625 N*m
#$&*
@& The displacement from x = .05 m to x = .20 m is .15 m, not .03 m.
Otherwise you're doing the right thing here.*@
What is the meaning of the altitude of this trapezoid?
****
amount of Newtons
#$&*
What is the meaning of the width of this trapezoid?
****
meters
#$&*
What therefore is the meaning of the area of this trapezoid?
****
amount of energy? (N*m)
#$&*
@& When you multiply force by displacement you get work.
If the force is the net force that work is equal to the change in kinetic energy.*@
`q012. For the rotation data you took in class:
What was the average rate of rotation in the trial where the added masses were at the end of the rotating beam?
****
1 rotation in 4 sec
.25 rotation in 1 sec
#$&*
What would then have been the initial rate of rotation (at the instant your finger lost contact with the system)?
****
0.0625?????
I have no idea
#$&*
@& You give the average rate of rotation as .25 rotations / second.
The system comes to rest.
If the rate of rotation changes at a constant rate, then
What is the average rate of rotation (your answer appears to be .25 rot / sec, which is fine)?
Since the system comes to rest at the end of the interval, what do you conclude must have been the initial rate?
Do your initial and final rates of rotation average out to the average rate you originally found?
*@
Assuming that those masses were 14 cm from the center of rotation, how fast were they moving, in cm / second, at that initial instant?
****
Circumference = 2 Pi 14 = 28 Pi / 4 sec = 7 Pi cm / sec
#$&*
@& Very good.
That would be the average speed, though, not the initial speed.
*@
Assuming that the masses were each 60 grams, what was the kinetic energy of each mass?
****
KE= 1/2 (60 kg) (7 Pi cm /sec)^2 = 1470 Pi g cm^2/sec^2
#$&*
@& You answered the last two questions in terms of the average, rather than the initial speed, but otherwise everything was correct.
*@
For University Physics Students only:
`q013. For the v vs. t trapezoid whose width is `dt and whose altitudes are v0 and vf:
What is the slope of the graph and what does the slope mean? Be sure to explain the entire interpretation of the slope.
****
#$&*
What is the area of the graph and what does the area mean? Be sure to explain the entire interpretation of the area.
****
#$&*
In terms of v0, vf, `dt, a and `ds what two equations do we get from the expressions for the slope and the area?
****
#$&*
What equation do we get when we eliminate vf from these two equations? Verify that you know how to do the algebra of this elimination.
****
#$&*
What equation do we get when we eliminate `dt from these two equations? Verify that you know how to do the algebra of this elimination.
****
#$&*
`q014. The following allow us to define work (F * `ds), kinetic energy (1/2 m v^2), impulse (F `dt) and momentum (m v):
If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `dt, what is the result? Show the
algebra of your solution, or verify that you can do the algebra easily.
****
#$&*
If we solve F_net = m a for a and plug the result into the second of the equations obtained above, then solve this equation for F `ds, what is the result? Show the
algebra of your solution, or verify that you can do the algebra easily.
****
#$&*"
@& You didn't get everything, but you got most of it. You're doing great here.
Check my notes.*@