What is the coefficient of friction between the block and the table?

Question

A block weighing 2 lb is forced against a horizontal spring of negligible mass, compressing the spring an amount x1 = 6 inches. Upon releasing the block, it moves on a horizontal table top a distance x2 = 2ft before coming to rest. The spring constant k is 8 lb/ft.

Answer 1

The potential energy of the spring must equal the work done to slide the block.

The potential energy of the spring E is ½ k x², where k = 8 lb/ft and x = 0.5 ft.

The work done to move the block is W = F x, where F is the frictional force and x is the distance the block moved, 2 ft. The friction force F is the normal force N multiplied by the coefficient of friction μ. Here N = 2 lb.

Equating Energy to Work,

½ (8 lb/ft) (0.5 ft )² = μ (2 lb) (2 ft)

μ = 0.25

Answer 2

(W/g)v^2 = kx^2 - 2μWx
v^2 = 2μg(s - x)
2μg(s - x)(W/g) = kx^2 - 2μWx
2μsW - 2μxW = kx^2 - 2μWx
μ = kx^2/(2s)
μ = 8*0.5^2/(2*2)
μ = 0.25