Question

You push a 3.2 kg block against a horizontal spring, compressing the spring by 16 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 62 cm from where you released it. The spring constant is 170 N/m. What is the coefficient of kinetic friction between the block and the table?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the block = m = 3.2 kg

Force constant of the spring = k = 170 N/m

Compression of the spring = X = 16 cm = 0.16 m

Coefficient of kinetic friction between the block and table =

Friction force on the block = f = mg

Distance traveled by the block from release point before coming to rest = L = 62 cm = 0.62 m

By conservation of energy the initial potential energy of the spring is converted into work done against friction as the block comes to rest again.

kX²/2 = fL

kX²/2 = mgL

(170)(0.16)²/2 = (3.2)(9.81)(0.62)

= 0.112

Coefficient of kinetic friction between the block and table = 0.112