# Can we find work done on this particle, and if not, what are we missing?

• arturo
In summary, Andrew thinks the homework statement involves calculating work done over time, but he is not sure if the given information allows him to do so.f

## Homework Statement

A force varies with time according to the expression F=aΔt, where a = 2.0 N/s.

From this information, can you determine the work done on a particle that experienced this force over a displacement of 0.50 m? W = F*d
Vf = Vo + aΔt
F = ma

## The Attempt at a Solution

First I looked for Work.
-W = F*d
-F = 2Δt
-d = .5
-W = Δt

Then I asked, can we find Δt?

This was my first try:
-Vf = Vo + aΔt
-Vf/a = Δt
-a = F/m
We don't know m (inertia), so that should be one answer.

Second try:
initial force must equal zero at t = 0
-Fi = 0
if we had Ff, we could divide it by 2 and get Δt.
So I thought that would be a second answer.

These two together were incorrect. I would assume I made a mistake on one of these methods or there is something else I am missing.

Thanks for any assistance,
arturo

#### Attachments

Hi arturo.
1. Note that the question asks you to "Check all that apply".
2. Since F = aΔt, a is something other than acceleration, I would change that "a" to something less confusing (eg. "x") if you are going to use "a" also for acceleration.
3. Re: velocity. Where does it say that the particle is initially at rest? Does it matter? Why? (hint: how is the force - which is a function only of time - but acting on the particle over a fixed distance in your frame of reference affected by the initial velocity of the particle in your frame of reference?)
4. Re: the final force. If you are going to determine the work, would you not have to know the force for all times?
5. Re: inertia. If you were given the time interval would you need the particle's mass in order to find the work done on the particle over the given distance?

AM

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• Bystander
It doesn’t really matter for answering this non-calculational question, but I thought I would point out that W = F d is only true if the force is constant. Since F is varying, here you would have to integrate.

Okay,
So I understand I cannot use Ff.

In response to Andrew~
3. Initial velocity impacts how fast you travel the distance, and then would impact the total time to travel, thus impacting force?
5. If you were given the time interval, would would work not just equal Δt ?

I believe there is a way to solve for Δt with mass and Vi using a = 2Δt/m and d = ViΔt + 1/2aΔt2. But I am not sure if that is the correct approach.

Thank you so much for your assistance.

3. Initial velocity impacts how fast you travel the distance, and then would impact the total time to travel, thus impacting force?
A slower start means less total displacement at any given time. So every portion of the path is covered later and, accordingly, with a higher force.
5. If you were given the time interval, would would work not just equal Δt ?
I am not sure that I understand what you are saying. Work and time are different units.
I believe there is a way to solve for Δt with mass and Vi using a = 2Δt/m and d = ViΔt + 1/2aΔt2
What matters is that the initial position, velocity and acceleration pattern over time completely determine the path of the object over time. The formulas you propose assume constant acceleration and will not do the trick.

Edit: If you have not been exposed to calculus then you have little chance of determining a formula for work done. If you have been exposed to calculus then you can integrate acceleration once to obtain velocity as a function of time, integrate velocity once to obtain displacement as a function of time, solve the resulting [cubic] polynomial for the time at which displacement hits the end point and then take the path integral of incremental work over time as the object moves between the start time and the end time.

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I see where I went wrong trying to use kinematics.

W = Δt was something I solved for
W = F*d
F = 2n/s * Δt
d = .5m
W = Δt joules?

Edit-
I think I get it, done a bit of calculus, I was approaching it wrong from the start. I’ll give it a shot now this way.
Thank you.

W = Δt joules?
That is still dimensionally inconsistent. The number of joules you get depends on the units you use to measure delta t.
W = Δt joules/second would at least be dimensionally consistent.

But it would still be wrong because W=F*d applies when force is constant. This force is not constant. You do realize that the question as posed does not require you to actually solve for work done, right?

That is still dimensionally inconsistent. The number of joules you get depends on the units you use to measure delta t.
W = Δt joules/second would at least be dimensionally consistent.

But it would still be wrong because W=F*d applies when force is constant. This force is not constant. You do realize that the question as posed does not require you to actually solve for work done, right?
Yeah I get that it doesn’t really need me to solve it. Got that that my previous method for work doesn’t make sense when force varies. If I solve for a with f=ma then integrate, the +C would be vo right? Then if I integrate again, the +c would be xo, which would be zero. I can solve that for t, set the bounds of my integral to zero and t, then integrate f over x? So the two things I would need would be vo and m.
Thanks for the assistance,
Arturo

Sounds like you are completely on the right track.
set the bounds of my integral to zero and t, then integrate f over x
For the final integral, you would be integrating f(t)dx/dt dt or, equivalently, f(t)v(t) dt.

Cool.

Thanks again! I feel like I have a much better understanding of work and using calculus in physics!
-arturo

• jbriggs444