Question

A cruise-ship passenger uses a shuffleboard cue to push a shuffleboard disk of mass 0.40 kg horizontally along the deck so that the disk leaves the cue with a speed of 8.5 m>s. The disk then slides a distance of 8.0 m before coming to rest.

Find the coefficient of kinetic friction between the disk and the deck.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the shuffleboard disk = m = 0.4 kg

Coefficient of kinetic friction between the disk and the deck =

Friction force on the shuffleboard disk = f

f = mg

Initial speed of the disk = V = 8.5 m/s

Distance traveled by the disk before coming to rest = D = 8 m

The kinetic energy of the disk is lost to the work done against friction.

mV²/2 = fD

mV²/2 = mgD

V²/2 = gD

(8.5)²/2 = (9.81)(8)

= 0.46

Coefficient of kinetic friction between the disk and the deck = 0.46