Q168 / 180 · KCET 2013 · Physics

A truck accelerates from speed $v$ to $2 v$. Work done in during this is

  • Correct answer
    A
    three times as the work done in accelerating it from rest to $v$
  • B
    same as the work done in accelerating it from rest to $\mathrm{V}$
  • C
    four times as the work done in accelerating it from rest to $v$
  • D
    less than the work done in accelerating it from rest to $v$

Solution

Work done to accelerate from 0 to $v$ is, $W_{1}=\frac{1}{2} m v^{2}$ Work done to accelerate from $v$ to $2 v$ is, $\begin{aligned} W_{2} &=\frac{1}{2} m(2 v)^{2} \\ \Delta W &=W_{2}-W_{1}=\frac{1}{2} m(2 v)^{2}-\frac{1}{2} m v^{2} \\ \Rightarrow \quad & 4 W_{1}-W_{1}=3 W_{1} \end{aligned}$

Hence, the correct answer is option: 'A'.

More Work Power Energy Questions