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My problem gives a time-dependent force as follows:

lets say that the force is fairly simple, $F=6t$

lets say that we want to find the work done in the 1st second.

Here's my approach:

$W=F(t).v$

So, for a small interval where force can be considered constant,

$dW=F.dv$

consult this

using the kinematical relations:

$v+dv=v+(6t/m)*dt$

$dv=6tdt$ (for m=1kg)

therefore,

$=>dw=6t*6tdt$

$=>dw=36t^2dt$

so we can integrate using any limits on time I guess? But in my case, this doesn't work (wrong answer) idk why, it seems correct to me lol

For anyone who wants the answer, its $4.5J$ (at least according to my textbook)

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  • $\begingroup$ find the area of the triangle of base length 1 and height 6. this is the change in momentum in the first second; which is also the total momentum of the particle since you state it starts at rest. Then multiply by $\frac{1}{m}$ for the velocity and compute the kinetic energy, $\frac{1}{2}mv^2$ which will equal Work done. there’s no need for integral calculus. $\endgroup$ – Ubaid Hassan Mar 24 '20 at 13:22
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You are confusing work and power.

Because of the pioneering work (no pun intended !!!) of James Watt, the unit of power is called the Watt and denoted by $W$. This should not be considered as the first letter of "work" in the physical meaning of the word. I think this may be the cause of your confusion.

You are supposed to compute the work.

Work is the integral in time of power.

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Wrong units

lets say that the force is fairly simple, $F=6t$

And that's your first problem, right here. You cannot disregard units and expect to get any correct or even well-defined result from your calculations.

The left-hand side is a force (in Newton), the right-hand side is a duration (in seconds). This cannot work, and you cannot use this equation anywhere.

You should replace it with $F = \frac{6\mathrm{N}}{\mathrm{s}} * t$

Back to basics

  • Time $t$ is in seconds.
  • Force $F$ is in Newtons.
  • Work is in Joules.
  • Power is in Watts.
  • Force * velocity is in Watts.
  • Force * distance is in Joules.
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  • $\begingroup$ Thats what i Thought, but the problem says just that. No extra info regarding the force itself. here's the problem if you want "A time-dependent force f=6t acts on a particle of mass 1kg. If the particle starts from rest, the work done by the force in the 1st second is...." $\endgroup$ – Shah Mohammad Arhum Dec 5 '19 at 9:03
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$$\begin{align}a &=\dfrac{F}{m} = \dfrac{6}{m}t \\ v(t)-v(0) &= \int\limits_0^t a~dt=?\end{align}$$

Using velocity you can find the kinetic energy(which is work done here if $v(0)=0$). See if you can finish it off..

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  • $\begingroup$ Beware ! Your answer is of course correct, but you are giving the entire answer to what is essentially a "homework and exercise" question. I have previously been censured by the "people in the know" for doing just that. We are only supposed to give hints, not solve the problem. $\endgroup$ – Alfred Dec 1 '19 at 6:28
  • $\begingroup$ Even a partial solution can be called "almost solution" if it gets too close. $\endgroup$ – Alfred Dec 1 '19 at 6:31
  • $\begingroup$ Ah ok. I'll remove some... just so you know I'm not so good with physics haha. Your method looks nice but I'm still wondering how to convert $F.dr$ to $P.dt$ $\endgroup$ – AgentS Dec 1 '19 at 6:31
  • $\begingroup$ Easy enough but this is not the place to tell you. I'll try to open a discussion. $\endgroup$ – Alfred Dec 1 '19 at 6:37
  • $\begingroup$ That would be awesome.. how do I open discussion? I don't see chat option on my end $\endgroup$ – AgentS Dec 1 '19 at 6:43

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