Question

A force acts on a 30g particle in such a way that the position of the particle as a function of time is given by x=3t-4t²+t³,where x is in metres and t is in seconds. The work done during the first 4 second is

Answer 1

Differentiate the equation wrt. time to find the expression for instantaneous velocity.

Find velocity (v1) at t=0

Find velocity (v2) at t=4s

Use work-energy theorem to find the work done, W=12mv22−12mv21

The answer is 528×$10^{-3}$J