91Ó°ÊÓ

The mass of the block, m = 2.0 kg

The spring constant of the spring k = 200 N/m

The compression of the spring x = 15 cm = 0.15 m

The distance travelled by the block d =75 cm = 0.75 m

The block is pushed against the spring and the spring gets compressed. As the block is released, the compressed spring pushes back the block. This converts the potential energy of the compressed spring to kinetic energy of the block. But since there is friction between the block and the tabletop, this kinetic energy is lost as thermal energy. Hence, the block comes to rest after travelling some distance. Thus, we need to apply the principle of conservation of energy to study the system.

Formula:

$$\begin{gathered} \mathrm{K}.\mathrm{E}=\frac{1}{2}m{v}^2 \\ \mathrm{P}.\mathrm{E}=\frac{1}{2}\mathrm{k}{\mathrm{x}}^2 \\ {E}_{th}=F_{f}d \end{gathered}$$

We use the principle of conservation of energy as

$$\begin{gathered} PE\left(spring\right)=KE\left(block\right)=E_{th} \\ {E}_{th}=PE\left(spring\right)=\frac{1}{2}k{x}^2 \\ {F}_{f}d=\frac{1}{2}×200×0.{15}^2 \\ =2.25 \\ {F}_{f}d=\mathrm{μ}mgd \\ \mathrm{μ}mgd=2.25 \\ \mathrm{μ}=\frac{2.25}{2.0×9.8×0.75} \\ =0.15 \end{gathered}$$

Hence, the block-table coefficient of friction is 0.15