# Various forces on various objects

## Homework Statement

sorry, me again.
The following figures show various forces acting on gliders of various masses at rest on a frictionless track. In each case, the force is applied for 30cm. The choice listing the examples in decreasing order for the increase in kinetic energy is {-4J < -2J < 1J} answer:(a) C > A=B > D (b) B=C > D > A (c) C > A > B > D (d) B > C > D > A (e) B > C > A > D

W=FD
F= MA

## The Attempt at a Solution

I dont clearly understand the questions here, is it asking for sorting it from most work to least? (cant ask teacher since he nevers check his email). if so wouldn't it be D>c>b>a since the mass of it all is combine on d, so it requires more force? I dont understand how in the answer, a block can be = to another block, of the other block (combined) mass is more?

## Answers and Replies

Doc Al
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What determines the change in kinetic energy?

CWatters
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I dont clearly understand the questions here, is it asking for sorting it from most work to least?

Yes. It's asking you to put them in order of the KE gained. So which one gains the most KE? By conservation of energy the KE gained = work done by the applied force.

if so wouldn't it be D>c>b>a since the mass of it all is combine on d...

No. Treat all of the drawings as separate experiments. They don't collide.

I dont understand how in the answer, a block can be = to another block

One can equal another if it gains the same KE as another.

What determines the change in kinetic energy?
mass and velocity, but how can I find velocity without the time? I'm looking at all the velocity equations I know and it all depend on time.

Doc Al
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mass and velocity, but how can I find velocity without the time? I'm looking at all the velocity equations I know and it all depend on time.
Forget all that. Think of the work-energy theorem.

Doc Al
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When I say "forget all that" I am being a bit facetious. You could do it that way, but that's way harder than necessary. You are given the force applied and the distance -- so use it.

When I say "forget all that" I am being a bit facetious. You could do it that way, but that's way harder than necessary. You are given the force applied and the distance -- so use it.
Yes. It's asking you to put them in order of the KE gained. So which one gains the most KE? By conservation of energy the KE gained = work done by the applied force.

No. Treat all of the drawings as separate experiments. They don't collide.

One can equal another if it gains the same KE as another.

Ok, so I calculated out the work done on each block seperately, A = 500J B=2000J C=1000J D= 1125J, so it's B>D>C>A, but that's not a choice, the closest is B=c>d>a and the reason why b=c is because c is half the weight as b, and if it was the same it would ACTUALLY be equal? am I correct?

edit: wait... why am I multiply force by the mass????????? ignore this then....

Doc Al
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Hint: The distance is the same for each, so just arrange by force.

• Just_enough
Hint: The distance is the same for each, so just arrange by force.
oh wow... that's so much easier

haruspex
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I'm looking at all the velocity equations I know and it all depend on time.
As Doc Al writes, you do not need to find the velocity. But if you insist on taking that route....
There are five standard variables in the equations for constant acceleration, the symbols for them spelling the acronym SUVAT.
Any four can be related by an equation, so there are five standard equations, each omitting one variable. In particular, one omits t. That one is effectively an energy equation, but with mass cancelled out.

oh wow... that's so much easier
what if say that it's being pushed for 1s? can we ignore 1s too since it's all being pushed for 1 s?

Doc Al
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what if say that it's being pushed for 1s? can we ignore 1s too since it's all being pushed for 1 s?
Nope. Since each will have a different acceleration, the distances will be different even though the time is the same. But in that case, you might want to use the impulse momentum theorem to help find the change in energy. (It will take a few steps.)

Nope. Since each will have a different acceleration, the distances will be different even though the time is the same. But in that case, you might want to use the impulse momentum theorem to help find the change in energy. (It will take a few steps.)
So I would have to find accel from the force, from there find the velocity of 1s, then from there find the impulse and sort the impulse in increasing order?
if so my increasing order should be B=C>D>A does that sound right?

Last edited:
CWatters
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Correct.
The easy way is to use KE=work=force*displacement.
The hard way would be to use Newton's law and the equations of motion (SUVAT) to calculate the final velocity for each. Then use KE=0.5mv2

Correct.
The easy way is to use KE=work=force*displacement.
The hard way would be to use Newton's law and the equations of motion (SUVAT) to calculate the final velocity for each. Then use KE=0.5mv2
if I were to use force*displaycement, what would happen to the time then? where would that go?

I'm confuse, still on the same model as original post, it's ask for " increasing order for the increase in momentum is" but it's giving me
(a) C < B < D < A (b) C < A=B < D (c) B=C < D < A (d) C < B=D < A (e) A < C < D < B and the closest to my answer is C, but the signs are wrong... I think.
Are the sign correct?

haruspex
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I'm confuse, still on the same model as original post, it's ask for " increasing order for the increase in momentum is" but it's giving me
(a) C < B < D < A (b) C < A=B < D (c) B=C < D < A (d) C < B=D < A (e) A < C < D < B and the closest to my answer is C, but the signs are wrong... I think.
Are the sign correct?
One of those answers is correct, but it is not C.

I've put the thread back to Unsolved.

One of those answers is correct, but it is not C.

I've put the thread back to Unsolved.
it's not? I thought it was list the impulse in increasing order

haruspex
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it's not? I thought it was list the impulse in increasing order
Sure, but you first have to calculate the impulses. I see no working by you in that regard.
I did not understand this part of an earlier post:
find accel from the force, from there find the velocity of 1s
Where do you get 1s from? In each case, the force is applied over the same distance. That distance will be covered in different times. So you cannot multiply the force by 1s to get the impulse.

Sure, but you first have to calculate the impulses. I see no working by you in that regard.
I did not understand this part of an earlier post:

Where do you get 1s from? In each case, the force is applied over the same distance. That distance will be covered in different times. So you cannot multiply the force by 1s to get the impulse.
from
So I would have to find accel from the force, from there find the velocity of [.1s (my bad)], then from there find the impulse and sort the impulse in increasing order?
if so my increasing order should be B=C>D>A does that sound right?

and I did the work on paper. I didnt multiply it by the seconds, I first found the accel by diving the force by mass, then from there use A=Δv/Δt for the interval of .1s to get velocity final, and from there I did mass time Δv (v final since initial is 0) which gave me the [change in momentum]

haruspex
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A=Δv/Δt for the interval of .1s
Using an arbitrary 0.1s is no more correct than using an arbitrary 1s.
The change in momentum is over whatever period the force is applied for. The greater the acceleration, the sooner the 30cm is covered, so the shorter the time.

Using an arbitrary 0.1s is no more correct than using an arbitrary 1s.
The change in momentum is over whatever period the force is applied for. The greater the acceleration, the sooner the 30cm is covered, so the shorter the time.
sorry, what? I dont understand what you mean

haruspex
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sorry, what? I dont understand what you mean
Explain how you justify multiplying the acceleration by 0.1s. Where does the 0.1 s come from?

Explain how you justify multiplying the acceleration by 0.1s. Where does the 0.1 s come from?
Oh, sorry, I was suppose to post the whole question, "If instead, the forces in the above figures are applied for 0.1s, the choice listing theexamples in increasing order for the increase in momentum is"

haruspex
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Oh, sorry, I was suppose to post the whole question,
That is generally a good idea.
"If instead, the forces in the above figures are applied for 0.1s, the choice listing the examples in increasing order for the increase in momentum is"
That makes a different one of the offered answers correct.
The method you described in post #20 is not the simplest, but it should have worked. Please post detailed working for at least one of the four arrangements.