Two skiers go to the top of a 50 meter high mountain. David (120 kg) takes the s

Question

Two skiers go to the top of a 50 meter high mountain. David (120 kg) takes the steep route while Teresa (60 kg) takes the less steep route. How fast will David be traveling? How fast will Teresa be traveling? Who will reach the bottom first? Assume energy is conserved. Two skiers go to the top of a 50 meter high mountain. David (120 kg) takes the steep route while Teresa (60 kg) takes the less steep route. How fast will David be traveling? How fast will Teresa be traveling? Who will reach the bottom first? Assume energy is conserved.

Explanation / Answer

Given,

mD = 120 kg ; mT = 60 kg ; h = 50 m

We know from conservation of energy

PE = KE

m g h = 1/2 m v^2

v = sqrt (2 g h)

since they have climbed the same height, both will have same potential energy. And since as stated in the question energy is conserved they will have same velocity irrespetive of taking less or more steep routes.

vD = vT = sqrt (2 x 9.81 x 50) = 31.32 m/s

vD = 31.32 m/s

vT = 31.32 m/s

v = D/t => t = D/v

since both of them have same velocity and the same ditance to be travelled, they will reach the bottom on same time.

[If there would be friction acting, then conservation of energy would not be applicable and all the above parameters would have been different for both of them].

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