A sled and rider of total mass 50 kg are perched at the top of a hill as shown.

Question

A sled and rider of total mass 50 kg are perched at the top of a hill as shown. The top of the hill is 60 m above the low point of the path of the sled. A hump in the hill is 20 m above this low point. Approximately 2000 J of work is done against friction as the sled travels between the start and the top of that hump, and there is no friction after that hump. (The snow has turned to ice there.)

a) If the sled starts with barely any push from the rider, how fast will it be going when it reaches the top of the hump?

b) If the sled starts with barely any push from the rider, how fast will it be going when it reaches the low point of the path?

c) Suppose the mass of the sled and rider is made twice as large, yet the 2000 J of work done against friction remains unchanged. Compared to your answer in part (b) above, the final speed of the sled would now be (Circle one:)

twice as big slightly bigger the same slightly less half as big

d) Compute the maximum height for the hump that will still permit the original 50-kg sled to just barely make it over that second hill.

Explanation / Answer

If we include the 'surroundings' as part of the total energy,that is E = mechanical energy + heat Height ofhill #1= h1 Height of hill #2 = h2 Reference level is at the ground At the top point: E1=KE+ PE +heat=PE = m gh1since it is not moving initially Top of 2nd hill: E2 =KE+ PE + heat= m g h2+2000J b/cit willjuststop at the top. Conservation:E1= E2 m g h1 = m g h2+2000 Solve for h2 m g h1 -2000 =m g h2 m g h1 -2000= h2 --------------- m g 50kg * 9.8 * 60 m - 2000 / (9.8 *50kg) =h2 = 55.91836734693878m

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