Question

You push a 2.0 kg block against a horizontal spring, compressing the spring by 15 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 75 cm from where you released it. The spring constant is 200N/m. What is the block-table coefficient of kinetic friction?

Short answer

The block-table coefficient of friction is 0.15

Step-by-step solution

 Step 1: Listing the given quantities

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

Understanding the concept of conservation of energy

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:

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

Calculation of coefficient of friction

We use the principle of conservation of energy as

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

Hence, the block-table coefficient of friction is 0.15