Chapter 5 - Work and Energy - Learning Path Questions and Exercises - Exercises - Page 176: 33

$200m$

Work done = Change in KE =$\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}=0-\frac{1}{2}mu^{2}=-\frac{1}{2}mu^{2}$ $-Force\times Distance=-\frac{1}{2}mu^{2}$ $Distance = \frac{m}{2F}u^{2}$ Hence, $\frac{S_{2}}{S_{1}}=(\frac{u_{2}}{u_{1}})^{2}$ $S_{2}=(\frac{90}{45})^{2}\times50m=200m$