Question

a force acts on a 4kg particle is in such away that the position of the particle as a function of time is given by x=6t-7t square what is the work done by the force during the first 3 second

Answer 1

Answer:

First differentiate the equation with respect to time and find the velocity. V= 3t-8t+3t^2 - (i) First time velocity ( V1)= 3 Now the time is given i.e., 4 second . Put in equation (i). We will get (V2)=19. Apply work energy theorem………………Work done = ∆k.E. ………………………………w= K.E final - K.E initial = 1/2 mv^2 - 1/2 mu^2. ..,…… 1/2 × 3o/1000×19 - 1/2× 30/1000×3………………solving this we will get 5.28J ( answer).