A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.39). When released, the block moves on a horizontal table top for 1.00 m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the table top?

Given Data:

The force constant of spring is:k=100N/m

The distance travelled by block before coming to rest is: l=1 m

The mass of block is: m =0.50 kg

The compression of spring is: x =0.20 m

The coefficient of kinetic friction is${\mu }_k$

The coefficient of kinetic friction is calculated by equating the elastic potential energy of spring in compression to work done by frictional force before coming to rest on the block.

The coefficient of kinetic friction between block and tabletop is given as:

$$\begin{gathered} \frac{1}{2}k{x}^2={\mu }_{k}mgl \\ {\mu }_k=\frac{k{x}^2}{2mgl} \end{gathered}$$

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

$$\begin{gathered} {\mu }_{\mathit{k}}=\frac{\left(100\mathrm{N}/\mathrm{m}\right){\left(0.20\mathrm{m}\right)}^2}{2\left(0.50\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(1\mathrm{m}\right)} \\ {\mu }_{\mathit{k}}=0.41 \end{gathered}$$

Therefore, the coefficient of kinetic friction between block and table top is 0.41.