Power and work contradiction

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 :)