A block of mass $m=1\text{ kg}$ is pushed against a spring with spring constant $k=100\text{ N/m}$, compressing it by $x=0.2\text{ m}$. When released, the block slides on a rough horizontal surface with coefficient of kinetic friction $\mu =0.2$. How far will the block travel before coming to rest?

Solution

1. The initial elastic potential energy stored in the spring is $$\begin{array}{rl}U& =\frac{1}{2}k{x}^2\\ & =\frac{1}{2}(100\text{ N/m})(0.2\text{ m}{)}^2\\ & =2\text{ J}\end{array}$$.
2. The work done by kinetic friction as the block travels a distance $d$ is $$\begin{array}{rl}{W}_f& =f_{k}d\\ & =\mu mgd\\ & =(0.2)(1\text{ kg})(10{\text{ m/s}}^2)d\\ & =2d\end{array}$$.
3. By the work-energy theorem, the initial energy equals the work done by friction: $2\text{ J}=2d$.
4. Solving for the distance gives $d=1.0\text{ m}$.
5. Hence the answer is (B).