Question

Under the action of a force a 2kg body moves such that the position x as a function of
time is given by x=2t^2, where x is in meters and t and is in seconds. The work done
by the force in the first three seconds is​

Answer 1

Answer:

144J

Explanation:

here position is dependent to time as

x=2t²

differenciating on both sides with respect to time

dx/d=4t

or,v=4t

at t=3 velocity v=4*3=12

the work done is given by change in KE

U=1/2*2*((12²-0)

   =144J