# Various forces on various objects

• Just_enough
In summary, the conversation is discussing a problem involving forces and kinetic energy. The task is to determine the order of increase in kinetic energy among different gliders on a frictionless track when a force is applied for a distance of 30cm. The answer is B=C>D>A, with the option of using either the work-energy theorem or the equations of motion to solve for the final velocity and kinetic energy.
Just_enough

## 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 don't 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 don't understand how in the answer, a block can be = to another block, of the other block (combined) mass is more?

What determines the change in kinetic energy?

Just_enough said:
I don't 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 don't understand how in the answer, a block can be = to another block

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

Doc Al said:
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.

Just_enough said:
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.

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.

Doc Al said:
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.
CWatters said:
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...

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

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

Just_enough said:
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 canceled out.

Just_enough said:
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?

Just_enough said:
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.)

Doc Al said:
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:
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

CWatters said:
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?

Just_enough said:
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.

haruspex said:
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

Just_enough said:
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:
Just_enough said:
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.

haruspex said:
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
Just_enough said:
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]

Just_enough said:
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.

haruspex said:
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 don't understand what you mean

Just_enough said:
sorry, what? I don't understand what you mean
Explain how you justify multiplying the acceleration by 0.1s. Where does the 0.1 s come from?

haruspex said:
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"

Just_enough said:
Oh, sorry, I was suppose to post the whole question,
That is generally a good idea.
Just_enough said:
"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.

haruspex said:
Please post detailed working for at least one of the four arrangements.
uh, do you mean show my work? if so, I just did like I say, find the accel of each the blocks, so for box a it's 50 = 10a then divide both side by 10 and go a= 5 (same for block b and D) and for block c I did 100/10 since the force was 100 and mass is 10 and got a=10.

from there, since block a,b, and d have accel of 5, I only need to find vf once, so I did 5=vf-vi/tf-ti, and vi and ti is 0, and what I got is 5=vf/.1s,

then times 5 by .1 to cancel out the denominator (and did this for acceleration of 10 for block c). after that, I multiply .5m/s to mass of blocks that have accel of 5 to get my momentum and multiply 1 by 10 for block c(A =5, B= 10, C =10, D = 7.5)

Just_enough said:
A =5, B= 10, C =10, D = 7.5
Which matches one of the choices, no?

haruspex said:
Which matches one of the choices, no?
The only choice on there with B=C is C, but you said it was in correct...

Just_enough said:
The only choice on there with B=C is C, but you said it was in correct...
Are we looking at the same list?
Just_enough said:
(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

haruspex said:
Are we looking at the same list?
no, you're looking at the original list instead of the new list I posted #16. (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

Just like when you were given that all the distances were the same and asked about the change in energy, there is an equally simple way to rank order the change in momentum when the times are all the same. Consider the impulse.

Doc Al said:
Just like when you were given that all the distances were the same and asked about the change in energy, there is an equally simple way to rank order the change in momentum when the times are all the same. Consider the impulse.
the difference in velocity? 3/4 objs have the same acceleration, so they end up with with the same velocity...

Just_enough said:
no, you're looking at the original list instead of the new list I posted #16. (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
In that case I agree with you: it is c but with the inequalities reversed.

Just_enough
Just_enough said:
the difference in velocity? 3/4 objs have the same acceleration, so they end up with with the same velocity...
But they have different masses.

But the same impulse gives the same change in momentum. (And if the times are all the same...)

## 1. What are the different types of forces that act on objects?

There are four fundamental forces that act on objects: gravity, electromagnetism, strong nuclear force, and weak nuclear force. Other common forces include friction, tension, and air resistance.

## 2. How do forces affect the motion of objects?

Forces can cause objects to accelerate, decelerate, or change direction. The direction and magnitude of the force determine the resulting motion of the object.

## 3. What factors influence the strength of a force?

The strength of a force depends on the mass and acceleration of the object. The greater the mass or acceleration, the stronger the force will be.

## 4. Can forces cancel each other out?

Yes, forces can cancel each other out if they are equal in magnitude and opposite in direction. This is known as balanced forces, and it results in no net force acting on the object.

## 5. How do forces affect the stability of an object?

Forces can either increase or decrease the stability of an object. A balanced force can help maintain the stability of an object, while an unbalanced force can cause it to become unstable and potentially fall over.

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