Various forces on various objects

[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

Homework Equations

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?

 

[Doc Al]

What determines the change in kinetic energy?

 

[CWatters]

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.

 

[Just_enough]

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]

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]

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.

 

[Just_enough]

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]

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]

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]

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]

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.)

 

[Just_enough]

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?

 

[CWatters]

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

 

[Just_enough]

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?

 

[Just_enough]

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]

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.

 

[Just_enough]

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]

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.

 

[Just_enough]

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]

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.

 

[Just_enough]

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

 

[haruspex]

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?

 

[Just_enough]

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]

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.

 

[Just_enough]

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 v_f once, so I did 5=v_f-v_i/t_f-t_i, and v_i and t_i is 0, and what I got is 5=v_f/.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)

 

[haruspex]

A =5, B= 10, C =10, D = 7.5

Which matches one of the choices, no?

 

[Just_enough]

Which matches one of the choices, no?

The only choice on there with B=C is C, but you said it was in correct...

 

[haruspex]

The only choice on there with B=C is C, but you said it was in correct...

Are we looking at the same list?

(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

 

[Just_enough]

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

 

[Doc Al]

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.

 

[Just_enough]

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...

 

[haruspex]

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.

 

[Doc Al]

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...)