A 73 kg skier coasts up a hill at 9.3 degrees to the horizontal. Determine how far along the...hill the skier slides before stopping, if the initial speed at the bottom is 4.2 m/s?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The kinetic energy that the skier has at the bottom of the hill is converted to gravitational potential energy as he moves up. The skier would stop when the kinetic energy has reduced to zero and it has been entirely converted to potential energy.

Kinetic energy is given by (1/2)*m*v^2. Using the information provided, it is (1/2)*73*(4.2)^2.

The potential energy is given by m*g*h, where m is the mass, g is the acceleration due to gravity and h is the height.

So we have (1/2)*73*(4.2)^2 = 73*9.8*h

=> (1/2)*(4.2)^2 = 9.8*h

=> h = (1/2)*(4.2)^2/9.8

As the slope is 9.3 degrees, the distance D moved up the slope is related to the height by h/D = sin 9.3

=> D = h/sin 9.3

=> D = [(1/2)*(4.2)^2/9.8]/sin 9.3

=> D = (1/2)*(4.2)^2/(9.8*sin 9.3)

solving for D we get,

=> D = 5.56 m

The skier can slide 5.56 m up the slope before he comes to a halt.

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