A body is starting from rest. A force is acting on it for a short period of time. In that given time, power delivered to it at any instance $$t$$ is given $$P = F \cdot v_1 = ma \cdot v_1 = mv_1^2/t,$$ where $$v_1$$ is the velocity of the body at that instant. However, $$P = \frac{\text{Work done}}{\text{time}} = \frac{\Delta (\text{Kinetic Energy})}{t} = \frac12 m (v_1^2-v_2^2),$$ where $$v_2$$ is the initial velocity. Since the body starts from rest, $$v_2 = 0$$. Thus, $$P = \frac12 m v_1^2 \frac{1}{t} = \frac{mv_1^2}{2t},$$

resulting in a contradiction. Can someone please explain where I went wrong?

• Are you familiar with the difference between average power and instantaneous power? Mar 1, 2021 at 14:35
• I am overriding the closure of this question as “homework-like.” I agree it is an elementary question, and there is probably a duplicate somewhere. It has the shape of an off-topic homework question if you only skim it, because the first sentence reads like a problem setup and the text before the question mark is “where have I gone wrong.” But the question here is not “what is the answer”; it is fundamentally “why are these two things different,” which is deeper and more interesting.
– rob
Mar 4, 2021 at 1:02

Can someone please explain where I went wrong?

You haven't been rigorous enough.

Newton's second law tells us: $$F=ma$$ $$a=\frac{F}{m}$$ Because the force applied is constant we can write: $$\Delta v=v_1-v_2=a\Delta t=\frac{F\Delta t}{m}$$ with: $$v_2=0$$ Calculating the work from the change in Kinetic energy $$K$$: $$W=\Delta K=\frac12 mv_1^2-\frac12 mv_2^2=\frac12 mv_1^2$$ $$P=\frac{1}{2\Delta t}mv_1^2$$ Again, because $$F$$ is constant: $$W=F\Delta x$$ $$P=\frac{\Delta W}{\Delta t}=F\frac{\Delta x}{\Delta t}=Fv_{av}$$

Here, we need to use the average speed $$v_{av}$$ because speed here varies (linearly) from $$0$$ to $$v_1$$, so: $$P=F\frac{v_1}{2}$$ $$P=ma\frac{v_1}{2}=m\frac{\Delta v}{\Delta t}\frac{v_1}{2}=\frac{v_1}{2\Delta t}mv_1=\frac{1}{2\Delta t}mv_1^2$$

So we get the same result both ways.

The equation $$P=Fv$$ (or more precisely $$P=\mathbf F\cdot\mathbf v$$) gives you the instantaneous power supplied by the force $$F$$.

The equation $$P=\Delta\text{KE}/\Delta t$$ is the average power supplied by the force over the time interval $$\Delta t$$.

This is why your answers are off by a factor of $$2$$. You are calculating two different things. The question asks for the power at any time, so you want the instantaneous one.

This goes like this $$P=\frac{dW}{dt}=\mathbf{F}\cdot\mathbf{v}=m\mathbf{a}\cdot \mathbf{v}$$ $$\mathbf{a}\cdot \mathbf{v}=\frac{d\mathbf{v}}{dt}\cdot \mathbf{v}=\frac{1}{2}\frac{d}{dt}v^2$$ So that $$\boxed{P=\frac{d}{dt}\left( \frac{1}{2}mv^2\right)}$$ As required :)

• Note that $$\frac{d}{dt}v^2=\frac{d}{dt}\mathbf{v}\cdot \mathbf{v}=\mathbf{v}\cdot \frac{d}{dt}\mathbf{v}+\frac{d\mathbf{v}}{dt}\cdot \mathbf{v}=2\frac{d\mathbf{v}}{dt}\cdot \mathbf{v}$$ Mar 1, 2021 at 13:30