# A skier of mass 55 kg slides down a sloe 11.7m long, inclined at an angle to the horizontal. The magnitude of the kinetic friction is 41.5 N. The skier's initial speed is 65 cm/s and the speed at...

A skier of mass 55 kg slides down a sloe 11.7m long, inclined at an angle to the horizontal. The magnitude of the kinetic friction is 41.5 N. The skier's initial speed is 65 cm/s and the speed at the bottom of the slope is 7.19 m/s. Determine the angle from the law of conservation of energy.

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### 2 Answers

A skier of mass 55 kg slides down a slope 11.7 m long, inclined at an angle to the horizontal. The magnitude of kinetic friction is 41.5 N and air resistance is negligible. The initial speed of the skier is 65 cm/s and the speed at the bottom of the slope is 7.19 m/s.

When the skier slides down the inclined slope, gravitational potential energy is converted to kinetic energy. If the angle of the slope made with the horizontal is x, the potential energy of the skier at the top is the slope is m*g*h = 55*9.8*11.7*sin x. The initial kinetic energy of the skier is (1/2)*55*(0.65)^2. At the bottom of the slope the gravitational potential energy is 0 and the kinetic energy is (1/2)*55*(7.19)^2. Also, 41.5 N is the magnitude of the kinetic friction; as the frictional force acts for 11.7 m, the work done by the frictional force is 11.7*41.5 J.

The law of conservation of energy can be used to give:

55*9.8*11.7*sin x + (1/2)*55*(0.65)^2 - 11.7*41.5 = (1/2)*55*(7.19)^2

=> sin x = 0.3

=> x = arc sin 0.3

=> x = 17.49 degrees.

The angle of slope of the incline is 17.49 degrees.

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