A light horizontal spring has a spring constant of 107 N/m. A 3.08 kg block is pressed against one end of the spring, compressing the spring 0.107 m. After the block is released, the block moves 0.226 m to the right before coming to rest. The acceleration of gravity is 9.81 m/s 2 . What is the coefficient of kinetic friction between the horizontal surface and the block

Given Information:

mass of block = m = 3.08 kg

spring constant = k = 107 N/m

compression length = x = 0.107 m

distance traveled after release = d = 0.226 m

Required Information:

coefficient of kinetic friction = μ = ?

coefficient of kinetic friction = 0.089

Explanation:

We know that work done against the friction is given by

W = Fd

W = μmgd

Where μ is the coefficient of kinetic friction, m is the mass of block and d is the distance traveled by block after it was released.

We know that potential energy stored in the spring is given by

U = ½kx²

Where k is the spring constant and x is the compression length

From the principle of work energy theorem,

W = U

μmgd = ½kx²

μ = kx²/2mgd

Put the given values into the above equation

μ = 107*0.107²/2*3.08*9.81*0.226

μ = 0.089

Therefore, the coefficient of kinetic friction between the horizontal surface and the block is 0.089.

see earlier answer