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:


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


consult this

using the kinematical relations:


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




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)

  • $\begingroup$ 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

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.


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.
  • $\begingroup$ 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

$$\begin{align}a &=\dfrac{F}{m} = \dfrac{6}{m}t \\ v(t)-v(0) &= \int\limits_0^t a~dt=?\end{align}$$

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

  • $\begingroup$ 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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