Question

A light horizontal spring has a spring constant of 105 N/m. A 2.00 kg block is pressed against one end of the spring, compressing the spring 0.100 m. After the block is released, the block moves 0.250 m to the right before coming to rest. What is the coefficient of kinetic friction between the horizontal surface and the block?

Answer 1

Answer:

The the coefficient of kinetic friction between the horizontal surface and the block is 0.536.

Explanation:

Given that,

Spring constant = 105 N/m

Mass of block = 2.00 kg

Compress = 0.100 m

Distance = 0.250 m

We need to calculate the coefficient of kinetic friction

Using relation between friction force and restoring force

$kx=\mu mg$

Put the value into the relation

$105\times0.1=\mu\times2.00\times9.8$

$\mu=\dfrac{105\times0.1}{2.00\times9.8}$

$\mu=0.536$

Hence, The the coefficient of kinetic friction between the horizontal surface and the block is 0.536.

Answer 2

Answer:

Coefficient of friction, $\mu=0.535$

Explanation:

Given that,

Spring constant of the spring, k = 105 N/m

Mass of the block, m = 2 kg

The compression in the spring, x = 0.1 m

Let $\mu$ is the coefficient of kinetic friction between the horizontal surface and the block. The frictional force is balanced by the restoring force in the spring such that,

$\mu mg=kx$

$\mu=\dfrac{kx}{mg}$

$\mu=\dfrac{105\ N/m\times 0.1\ m}{2\ kg\times 9.8\ m/s^2}$

$\mu=0.535$

So, the coefficient of kinetic friction between the horizontal surface and the block is 0.535. Hence, this is the required solution.