If your solution to stated problem does not match the given solution, you should self-critique per instructions at

 

   http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution.  If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution. 

014.  `query 14

Question:  `qset 3 intro prob sets

If you calculate the acceleration on a mass m which starts from rest under the influence of a constant net force Fnet and multiply by a time interval `dt what do you get? 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a The acceleration of the mass is a = F_net / m, so the velocity of the object changes by amount

Since the initial velocity is zero, this will also be the final velocity:

From this and the fact that acceleration is constant (const. net force on const. mass implies const. acceleration), we conclude that

Multiplying this by the time interval `dt we have

If we multiply this by F_net we obtain

From our earlier result vf = F_net / m * `dt we see that

Since we started from rest, the final KE of the mass on this interval is equal to the change in KE on the interval.

We call F_net * `ds the work done by the net force.  Our result therefore confirms the work-kinetic energy theorem:

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Question:  `q Define the relationship between the work done by a system against nonconservative forces, the work done against conservative forces and the change in the KE of the system.  How does PE come into this relationship?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** The system does positive work at the expense of its kinetic and/or potential energy.

The work done by the system against all forces is `dW_net_BY.

`dW_net_BY is equal and opposite to `dW_net_ON, which is in turn equal to `dKE, the change in the kinetic energy of the system.

We conclude that `dW_net_BY = - `dKE.    The change in KE is equal and opposite to the work done by the system against the net force acting on it.

To consider the role of PE, we first review our formulation in terms of the work done ON the system:

`dW_net_ON = `dKE.

The work `dW_net_ON is the sum of the work done on the system by conservative and nonconservative forces:

and `dW_cons_ON is equal and opposite to `dPE, the change in the system's PE.

Thus `dW_net_ON = `dW_NC_ON - `dPE so that `dW_net_ON = `dW_cons_ON + `dW_NC_ON becomes

Since `dW_NC_BY = - `dW_NC_ON, we see that

Intuitively, if the system does positive work against nonconservative forces, `dPE + `dKE must be negative, so the total mechanical energy PE + KE of the system decreases.  (Similarly, if the system does negative work against nonconservative forces that means nonconservative forces are doing positive work on it, and its total mechanical will increase).

As usual, you should think back to the most basic examples in order to understand all these confusing symbols and subscripts (e.g., if I lift a mass, which you know intuitively increases its gravitational potential energy, I do positive work ON the system consisting of the mass, the conservative force of gravity acts in the direction opposite motion thereby doing negative work ON the system, and the work done BY the system against gravity (being equal and opposite to the work done ON the system by gravity) is therefore positive).

The equation -`dW_NC_BY = `dPE + `dKE isolates the work done by the system against nonconservative forces from the work it does against conservative forces, the latter giving rise to the term `dPE.

If the system does positive work against conservative forces (e.g., gravity), then its PE increases.

If the system does positive work against nonconservative forces (e.g., friction) then `dPE + `dKE is negative:  PE might increase or decrease, KE might increase or decrease, but in any even the total PE + KE must decrease.  The work done against a nonconservative force is done at the expense of at least one, and maybe both, the PE and KE of the system.  (In terms of gravitational forces, the system gets lower or slows down, and maybe both, in order to do the work).

If nonconservative forces do positive work on the system, then the system does negative work against those forces, and `dW_NC_ON is negative.  Thus -`dW_NC_ON is positive, and  `dPE + `dKE is positive.  Positive work done on the system increases one or both the PE and the KE, with a net increase in the total of the two.  (In terms of gravitational forces, the work done on system causes it to get higher or speed up, and maybe both.)

STUDENT RESPONSE WITH INSTRUCTOR COMMENTARY:  The work done by a system against nonconservative forces is the work done to overcome friction in a system- which means energy is dissipated in the form of thermal energy into the 'atmosphere.'

 

Good.  Friction is a nonconservative force. 

 

However there are other nonconservative forces--e.g., you could be exerting a force on the system using your muscles, and that force could be helping or hindering the system.  A rocket engine would also be exerting a nonconservative force, as would just about any engine.  These forces would be nonconservative since once the work is done it can't be recovered.

 

STUDENT RESPONSE WITH INSTRUCTOR COMMENTS:  The work done by a system against conservative forces is like the work to overcome the mass being pulled by gravity.

 

INSTRUCTOR COMMENT:  that is one example; another might be work to compress a spring 

STUDENT QUESTION

ok, alot to absorb but I think I am getting there. So KE is equal to work done ON the system not BY the system...this
is a little confusing. I was thinking that when we calculate the dw= fnet * 'ds we were caculated the work done BY the system not on the system.?

INSTRUCTOR RESPONSE

To be very specific

is the work done by the net force acting ON the system, where F_net_ON is the net force acting on the system.

The work-kinetic energy theorem states that

If positive work is done on a system, it speeds up. If negative work is done on the system, it slows down.

From the point of view of the system, if positive work is done by the system then the system has to 'use up' some of its kinetic energy to do the work, so it slows.

Positive work done BY the system constitutes negative work being done ON the system.

If part of the net force is conservative, then `dW_net_ON can be split into `dW_net_ON_cons and `dW_net_ON_noncons.

The quantity `dPE, the change in PE, is defined to be equal and opposite to `dW_net_ON_cons. That is,

It follows that `dW_net_ON = `dW_net_ON_noncons - `dPE, so that the work-kinetic energy theorem can be rewritten as

This is commonly rearranged to the form

STUDENT COMMENT (confused by too many symbols)

Once again this makes no since to me. All the symbols lost me

<h3>We can say this first in words, then translate the words into symbols:

remember that

Now, some forces are conservative and some are nonconservative, so

Since work on system by all forces = change in kinetic energy

Saying exactly the same thing in symbols:

Some forces are conservative and some are nonconservative, so

Since `dW_net_on = `dKE


STUDENT QUESTION

I do not understand conservative and nonconservative forces at all. Could you explain this to me in simpler terms?

INSTRUCTOR RESPONSE


The official definition is that conservative forces are path-independent. However that in itself is a tough concept to understand and I don't find it very useful when students first encounter the idea.

The basic idea itself is simple enough. A conservative force 'stores up' the work you do against it. So for example when you lift something you do work against gravity. If you then release it, gravity does equal work on it as it falls back to its original position.

As another example suppose you hang an object from a rubber band and allow it to come to rest at an equilibrium position. If you then pull it further downward you do work against the increasing tension in the rubber band. If you release it, the tension does work on the object as the rubber band springs back. The rubber band 'stored up' the energy you put into it when you pulled the object down, and returned the energy when you released the object.

The energy that was 'stored up' is called potential energy. When the object was released the potential energy was converted to kinetic energy.

The gravitational force is completely conservative. All the work you did against gravity to lift the object is returned. Gravity exerts just as much force on the return as it did when

The elastic tension force of the rubber band is not completely conservative. The rubber band heats up when it is stretched (and cools when it snaps back), with the net result being that some of the work you do goes into thermal energy (i.e., heat) so the rubber band doesn't manage to 'store' all the energy you put into it and you don't get all your energy back. So the rubber band force can be regarded as 'partially conservative'. It stores energy, but not all of it.

Friction is an example of a force that isn't conservative at all. If you push a box across the floor, it's friction that resists your efforts. However friction doesn't 'store' any of the energy you put into pushing the box. If you push the box from this side of the room to that side, then release it, the box doesn't slide back even a little bit. The energy you expended is simply dissipated (friction heats up the floor and the box, and the resulting thermal energy is just dissipated to the surroundings).



*@</h3>

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Question:  `qclass notes:  rubber band and rail

[ this refers to the activity in which we take one of the metal ramps, which we call a 'rail', and accelerate it across the tabletop with a rubber-band slingshot; the 'rail'
slides across the tabletop and comes to rest under the influence of friction   we can measure the frictional force and the force vs. length characteristics of the rubber band, and consider energy conservation with respect to this system ]

How does the work done to stretch the rubber band compare to the work done by the rubber band on the rail, and how does the latter compare to the work done by the rail against friction from release of the rubber band to the rail coming to rest?

 

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** The work done to stretch the rubber band would in an ideal situation be available when the rubber band is released.

 

Assuming that the only forces acting on the rail are friction and the force exerted by the rubber band, the work done by the rail against friction, up through the instant the rail stops, will equal the work done by the rubber band on the rail.

 

Note that in reality there is some heating and cooling of the rubber band, so some of the energy gets lost and the rubber band ends up doing less work on the rail than the work required to stretch it.  **

 

 

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Question:  `qWhy should the distance traveled by the rail be proportional to the F * `ds total for the rubber band?

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** Bottom line: 

F_net_ave * `ds between the initial state of rest and max KE must therefore be equal and opposite to F_net_ave * `ds between max KE and the final state of rest. 

During the second interval, therefore, F_net_ave remains constant, so it is the coasting displacement that varies with pullback.  The coasting displacement is therefore proportional to the F * `ds total for work done by the rubber band on the system.

More details:

The F_`ds total for the rubber band is the work done to accelerate the rail to its maximum velocity v_max.

Let's denote this simply by F_ave * `ds, where F_ave is understood to be the average force exerted by the rubber band (the rubber band force is at its maximum when the rubber band is pulled back, and decreases to 0 as it 'snaps back', accelerating the rail; so it makes sense to talk about the average rubber band force) and `ds is the displacement through which this force acts (i.e., the displacement from release until the rubber band loses contact with the rail).

While in contact with the 'rail' the rubber band exerts its force in the direction of the system's motion and therefore does positive work.  So F_ave * `ds is positive.

The 'rail' then coasts to rest subject to the force of friction, which acts in the direction opposite motion and therefore does negative work.  Assuming the frictional force f_frict to be constant, and using `ds_coast for the coasting displacement, the work done against friction is f_frict * `ds_coast.

For simplicity of notation we will neglect the presence of the frictional force during the first interval, while the rubber band is in contact with the 'rail'.  It isn't completely accurate to do so, but if the displacement during this interval is small compared to the coasting distance the error is small.  A comment at the end will indicate how to easily modify these results.

We will also neglect any other forces that might be acting on the system, so that the net force for the first phase is just the rubber band force, and for the second phase the net force is just the frictional force.

Now, during the first interval the rail's KE changes from 0 to 1/2 m v_max^2, where m is its mass, so by the work-KE theorem

During the second interval the rail's KE changes from 1/2 m v_max^2 to 0, so that

Thus F_ave * `ds = - f_frict * `ds_coast so that the coasting displacement is

F_ave and f_friction are in opposite directions, so if F_ave is positive f_frict is negative, making -1 / f_frict negative and

indicates a direct proportionality between `ds_coast and F_ave * `ds.

The above relationship tells us that the coasting displacement is proportional to the F * `ds total for the force exerted by the rubber band.

To correct the oversimplification of the given solution, if that oversimplification bothers you, you may proceed as follows (however if you find you don't completely understand the preceding you shouldn't confuse yourself with this until you do):

SYNOPSIS

The 'rail' is brought to rest by a pretty-much-constant frictional force. The work it does against this force is therefore proportional to how far it slides.

The 'rail' comes to rest when the frictional force has dissipated its initial kinetic energy, which was obtained from the rubber band.

So initial KE, which is pretty much equal to F * `ds for the rubber band, is proportional to how far the 'rail' slides.

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Question (Openstax) : 

`prin, `gen:  Consider the baby being weighed in Figure 4.34.

 (a) What is the mass of the child and basket if a scale reading of 55 N is observed?

(b) What is the tension T1 in the cord attaching the baby to the scale?

(c) What is the tension T2 in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg?

(d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.

Baby supported from scale, scale in turn supported from ceiling.

Figure 4.34 A baby is weighed using a spring scale.

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:

Given Solution: 

The weight of the baby is the force exerted on it by gravity.  That force is the product m * g of the baby's mass and the acceleration of gravity.  Thus

m * g = 55 Newtons  so that

m = 55 Newtons / g = 55 kg m/s^2 / (9.8 m/s^2) = 5.6 kilograms.

The baby is not accelerating so the net force on it is zero.  The only forces acting on it are the 55 Newton gravitational force, which acts downward, and the tension in the cord, which acts upward. Your figure should show a downward force of 55 Newtons and the upward force T_1.

Choosing upward as the positive direction, then, we have

T_1 - 55 Newtons = 0  so that

T_1 = 55 Newtons.

The net force on the scale is also zero, since it is not accelerating.  The tension in cord 1 acts downward on the scale.  So the forces on the scale are -T_1, T_2 and the gravitational force -0.50 kg * 9.8 m/s^2 = -4.9 Newtons.  Your figure should show the downward force -T_1, the downward for -4.9 N and the upward force T_2.

Thus we have

-T_1 + T_2 + -4.9 Newtons = 0   so that

T_2 = T_1 + 4.9 Newtons = 55 Newtons + 4.9 Newtons = 59.5 Newtons.

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Question:  `qgen phy A person of mass 66 kg crouches then jumps to a height of .8 meters.  From the crouched position to the point where the person leaves the ground the distance is 20 cm.  What average force is exerted over this 20-cm distance?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** Bascially the person rises 1 meter from the crouched position.

This requires a change in gravitational PE, which is easily calculated.

This change requires that work be done, and work can only be done while the player's feet are on the floor. Afterwards they are not in contact with anything and no upward force can be exerted.

The upward force is exerted by the floor in reaction to downward force exerted by the player's legs.

The upward force is exerted while the player moves 20 cm.

Having calculated the required PE change you know how much work must be done during the 20 cm interval. So you know the displacement and need only find the average force.

More formally, for the interval from the floor to the highest point of the jump:

`dW_noncons_ON = F_floor * 0.2 m
`dPE = 1 m * 66 kg * 9.8 m/s^2
`dKE = 0 (for the interval from crouched position to highest point the player is at rest at the initial and the final instant).

`dW_nc_ON = `dPE + `dKE so

F_floor * 0.2 m = 1 m * 66 kg * 9.8 m/s^2 + 0

F_floor = 1 m * 66 kg * 9.8 m/s^2 / 0.2 m = 3200 N, approx..

 

A quick solution:  This net force must do enough work to raise the person's weight 1 meter while acting through only a .2 meter displacement, and must therefore be 5 times the person's weight.  The normal force is equal to the net force plus the person's weight, so is 6 times the person's weight.

 

The detailed reasoning of another equivalent explanation is as follows:

`dW_nc_ON = `dPE + `dKE. 

`dKE is zero, since the person is at rest before the jump and at the top of the jump.  So the work done by the nonconservative force exerted by the floor must be equal to `dPE.

At the top of the jump the mass is 1 meter higher than at the bottom, so gravitational PE has increased by

Thus `dPE = +650 J.

The PE increase is due to the work done by the normal force (exerted by the floor) during the .2 meter interval before leaving the floor.  Thus

 

 

so that

An average force of 3300 N is required to make this jump from the given crouch. 

 

The information given in this problem probably doesn't correspond with reality.  A 3300 N force is equivalent to the force exerted by a 250-lb weightlifter doing a 'squat' exercise with about 500 pounds on his shoulders.  It is unlikely that a 66 kg player would exert this much aveage force throughout a jump (the force during the jump will be variable, and the maximum force might well exceed 3300 N; however as the player accelerates the force will likely decrease). 

 

More likely the 'crouch' required for a 1-meter jump would be significantly more than 20 cm.  A 20-cm crouch is only about 8 inches and vertical jumps typically involve considerably more crouch than this.  With a 40-cm crouch such a jump would require only half this force calculated here, and is probably feasible.  **

 

STUDENT QUESTION

 I was not sure how to find the average force so I just multiplied the mass by 9.8. After looking at the solution I am still confused on finding the normal force of the object.

INSTRUCTOR RESPONSE

Imagine you are lying on the floor, with 2' x 2' piece of plywood resting on your chest, and a child about half your weight standing on the plywood. If the child does a vertical leap, what will that feel like to you?

You can imagine the same with a person of your own size. If it's an average person with, say, a 20-inch vertical leap, what would that feel like? If it's someone with a 40-inch vertical leap, what would that feel like?

In each case you are experiencing the normal force exerted by the person on the board, which in turn exerts this force (plus a little bit more, due to the weight of the board itself) on you.  You in turn exert an equal and opposite upward force, which is what causes the person to rise. When the person is just standing on the board, the normal force is just equal and opposite the person's weight (we will regard the weight of the board as negligible) and it's not difficult for you to manage. When the person jumps, his or her legs push down, hard, and the board (and you by extension) have to push back. In the case of the 40-inch leap, you'll probably find that to be at least uncomfortable.

For the person to leap the floor has to push up hard enough to result in a net force equal to mass * acceleration (F_net = m a). Gravity is pulling down, so the net force is (net force) = (floor push) - (weight), so (floor push) = (net force) + (weight).

STUDENT QUESTION

this has .2 I am not sure were .2 came from, but in this problem we have .8

INSTRUCTOR RESPONSE

The person starts from a .2 meter crouch.

The player's altitude changes by `dy = 1 meter, from the .2 meter crouch to the .8 meter height. So PE increases by `dPE = weight * `dy. In this case `dPE is about 650 Joules.

The player's feet stay in contact with the floor for the first .2 meters. Only during this interval is force being exerted between feet and floor. So the work required to increase the PE is done during the .2 meter displacement.

Thus F_net * .2 meters = 650 Joules and F_net = 650 J / (.2 m) = 3250 N, approx..

F_net = F_normal + weight, so F_normal, the force between feet and floor, is F_net + weight. This comes out around 3900 N.

SYNOPSIS

It takes about 650 Joules to raise a 650 N person to a height of 1 meter.

The person cannot exert any lifting force after leaving the floor, which happens after traveling .20 meters.

So the person has to do 650 Joules of work while traveling .20 meters.

This requires 3300 N of force.

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Question (Openstax):   `gen Calculate the force a 70.0-kg high jumper must exert on the ground to produce an upward acceleration 4.00 times the acceleration due to gravity.

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating:

Given Solution: 

The acceleration of gravity is 9.8 m/s^2, so four times this acceleration is 4 * 9.8 m/s^2 = 39.2 m/s^2.

To produce this acceleration the net force on the jumper must be equal to m * a, where m is the 70 kg mass and a is the required 39.2 m/s^2 acceleration.

The forces on the jumper are the force exerted by gravity (i.e., the jumper's weight) and the upward force exerted by the floor.  To produce the required upward force the jumper must exert an equal and opposite downward force on the floor.

If F_floor is the force exerted by the floor, then the net force on the jumper is

F_net = F_floor - m * g

where we have chosen the upward direction as positive.

It follows that

F_floor = F_net + m * g = 70 kg * 39.2 m/s^2 + 70 kg * 9.8 m/s^2 = 3470 Newtons.

STUDENT QUESTION

 According to the equation Fnet=m*a (70kg*39.2m/s^2 =2744N) it takes 2744N to accelerate a mass of 70 kg to 39.2m/s^2.
So why do we have to add the weight of the jumper as well? I thought that was included in the original equation?



INSTRUCTOR RESPONSE


The net force is equal to the mass * acceleration.

The net force is equal to the sum of the forces acting on the jumper.

The normal force and the force of gravity both act on the jumper.

So

Gravitational force is -70 kg * 9.8 m/s^2 = -700 N, approx..

Thus

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Question:  `quniv phy  text prob 4.46 (11th edition 4.42) (4.40 in 10th edition) Mercury lander near surface upward thrust 25 kN slows at rate 1.2 m/s^2; 10 kN speeds up at .8 m/s^2; what is weight at surface?

 

 

Your solution: 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence rating::

Given Solution: 

`a** If a landing craft slows then its acceleration is in the direction opposite to its motion, therefore upward.

 

If it speeds up while landing that its acceleration is in the direction of its motion, therefore downward.

 

If the upward motion is taken as the positive direction, then the acceleration under a thrust of 25 kN is + 1.2 m/s^2, and the acceleration when under thrust of 10 kN is - .8 m/s^2.

 

In either case m * a = net force. 

Net force is thrust force + gravitational force.

We will take the downward direction to be positive.  This will give us a positive result for the weight m g.

1 st case, net force is m g - 25 kN so m * 1.2 m/s/s = m g - 25 kN.

1 st case, net force is 10 kN so m * (.8 m/s/s ) = m g - 10 kN.

 

Solve these equations simultaneously to get the weight m * g (multiply 1 st eqn by 2 and 2d by 3 and add equations to eliminate the first term on the left-hand side of each equation; solve for m * g). 

 

The solution is m * g = 16 kN.

 

Another solution:

 

In both cases F_net / a = m so if upward is positive and weight is wt we have

 

(25 kN - wt) / (1.2 m/s^2) = m and

(10 kN - wt) / (-.8 m/s^2) = m so

 

(25 kN - wt) / (1.2 m/s^2) = (10 kN - wt) / (-.8 m/s^2).

 

Solving for wt we get 16 kN. **

 

ANOTHER NEARLY IDENTICAL SOLUTION, WHICH ALSO SOLVES FOR THE ACCELERATION OF GRAVITY ON MERCURY

F_net = m a.

Choose upward as positive.

At 25 000 N we have
F_net = m * 1.2 m/s^2,

and

at 10 000 N we have
F_net = m * (-.8 m/s^2).

In the first case

F_net = 25 000 N - m g_planet

and in the second

F_net = 10 000 N - m g_planet.

So

25 000 N - m g_planet = m * 1.2 m/s^2

and

10 000 N - m g_planet = m * (-.8 m/s^2).

We can eliminate the unknown g_planet by subtracting the second equation from the first, to get

15 000 N = m * 2 m/s^2

so that

m = 15 000 N / (2 m/s^2) = 7500 kg.

Then the first equation becomes

25 000 N = 7500 kg ( 1.2 m/s^2 + g_planet)

so that

7500 kg * g_planet = 16 000 N

and

g_planet = 2.13 m/s^2.

So the weight of the lander is

7500 N * 2.13 m/s^2 = 16 000 N.

The net force with 25 000 N thrust is 9 000, resulting in acceleration 9 000 N / (7500 kg) = 1.2 m/s^2.

With 10 000 N thrust the resulting acceleration is -6000 N / (7500 kg) = -.8 m/s^2.

 

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