Question

A truck (6500 kg) is accelerating from 20.0 m/s to 30.0 m/s. calculate the work done?

Answer 1

Answer

Average acceleration in time t

$a=\frac{30-20}{t}=\frac{10}{t}m/s^2$

Average distance travelled

$d=(30-20) t$

Work done

$W=6500\times \frac{10}{t}\times10t=650KJ$