# A 7kg box starts 8m above the floor while moving 3m/s. It then slides down a ramp to a rough, level floor and comes to a stop in 12m. What is the coefficient of friction?

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A 7 kg box starts moving 8 m above the floor at 3 m/s. It slides down a ramp and comes to a rough level floor where it comes to a halt after traveling 12 m. It is assumed that the ramp is frictionless.

The potential energy of the box when it is at a height of 8 m is 7*9.8*8 = 548.8 J. This is converted to kinetic energy when the box slides down. Its initial speed was 3 m/s. At the bottom of the ramp its speed is 3 + sqrt(548.8*2/7) = 15.52 m/s

Let the coefficient of friction of the level floor be denoted by C. The normal force of the box is 7*9.8 = 68.6 N. The frictional force is 68.6*C. The deceleration due to this is 9.8C. As the box comes to a halt in 12 m

0 - 15.52^2 = 2*9.8*C*12

=> C = 1.024

**The coefficient of friction is 1.024**