# 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)

• 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. – 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.
• 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...." – 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..

• 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. – Alfred Dec 1 '19 at 6:28
• Even a partial solution can be called "almost solution" if it gets too close. – Alfred Dec 1 '19 at 6:31
• 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 $P.dt$ – AgentS Dec 1 '19 at 6:31
• Easy enough but this is not the place to tell you. I'll try to open a discussion. – Alfred Dec 1 '19 at 6:37
• That would be awesome.. how do I open discussion? I don't see chat option on my end – AgentS Dec 1 '19 at 6:43