Question

Physics. Work Problem?

A single force acts on a 2.5 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.8t - 1.7t2 + 0.95t3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.9 s.
my answer was 488.72 J but it was wrong.

Answer 1

Work done by an external force = change in kinetic energy.

Given,`x=3.8t-1.7t^2+0.95t^3`

So,`v=(dx)/(dt)=3.8-3.4t+2.85t^2`

So,using this equation,we get,at `t=0,v_o=3.8ms^-1`

And at `t=8.9,v_t=199.3 ms^-1`

So,change in kinetic energy =`1/2*m*(v_t^2 - v_o^2)`

Putting the given values we get,`W=K.E=49632.55J`