# 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.

Mar 15, 2018

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

Given,$x = 3.8 t - 1.7 {t}^{2} + 0.95 {t}^{3}$

So,$v = \frac{\mathrm{dx}}{\mathrm{dt}} = 3.8 - 3.4 t + 2.85 {t}^{2}$

So,using this equation,we get,at $t = 0 , {v}_{o} = 3.8 m {s}^{-} 1$

And at $t = 8.9 , {v}_{t} = 199.3 m {s}^{-} 1$

So,change in kinetic energy =$\frac{1}{2} \cdot m \cdot \left({v}_{t}^{2} - {v}_{o}^{2}\right)$

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