Work Done by a time-variable Force 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,
$dW=F.dv$

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)
 A: 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. 
A: 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.

A: $$\begin{align}a &=\dfrac{F}{m} = \dfrac{6}{m}t \\ v(t)-v(0) &= \int\limits_0^t a~dt=?\end{align}$$
Using velocity you can find the kinetic energy(which is work done here if $v(0)=0$). See if you can finish it off..
A: $F = 6t \implies a = \frac{dv}{dt} = 6t/m = 6t$
$\implies v = 3t^{2}$
$\implies dx = 3t^{2}dt$  as  $v = \frac{dx}{dt}$
$W = \int Fdx = \int 6t\cdot 3t^{2} dt = \frac{9}{2}t^{4}$
Therefore, at $t = 1s, W = 4.5 J$
