Work Done by a General Variable Force — Question 9

Question 9

A force-time area is not work without the motion

A 2 kg particle starts from rest at t=0. Its only force along x is $F_x=4t$ N, with t in seconds, for $0\le t\le2$ s.

1. Find velocity as a function of time.

2. Find work from $\int F_xv_x\,dt$.

3. Check the work using kinetic energy and identify the units of $\int F_x\,dt$.

Question 9 - Solution

A force-time area is not work without the motion

(a) Integrate acceleration first.

The governing relation is $$\boxed{v(t)=v_0+\frac1m\int_0^tF(\tau)\,d\tau,\qquad W=\int Fv\,dt}$$ Integrate the acceleration with $v(0)=0$: $$a(t)=\frac{(4\ \mathrm{N/s})t}{2\ \mathrm{kg}}=(2\ \mathrm{m/s^3})t,$$ $$\boxed{v(t)=\int_0^ta(\tau)\,d\tau=(1\ \mathrm{m/s^3})t^2}.$$

(b) Use displacement per unit time.

$$W=\int_0^2(4t)(t^2)\,dt=[t^4]_0^2=16\ \mathrm J.$$

(c) Perform a separate endpoint check.

At $t=2$ s the speed is 4 m/s, so the independent energy check gives $$\boxed{\Delta K=\tfrac12(2)(4)^2=16\ \mathrm J}.$$ The force–time integral is instead $$\int_0^2F_x\,dt=8\ \mathrm{N\,s},\qquad[\mathrm{N\,s}]=[\text{momentum}]\ne[\text{energy}].$$ It measures impulse; the additional velocity factor is necessary for work.