You are teaching your 30 kg younger sibling to ice skate at the bottom of a 5 m

Question

You are teaching your 30 kg younger sibling to ice skate at the bottom of a 5 m high icy hill. A 70 kg person at the top of the hill starts from rest and slides down the hill on their skates. Seeing this, you decide to push your sibling directly at the person with a velocity of 2 m/s such that the two collide at the very bottom of the hill. 5 m from the bottom of the hill is a patch of grass. If your sibling holds onto the person for dear life and they hit the patch of grass which has a coefficient of kinetic friction between the grass and the skates of 0.6, how far did the two travel from the base of the hill?

Explanation / Answer

Kinetic energy at the bottom of hill = (0.5 * 30 * 2^2) + (70 * 9.8 * 5) = 3490 J This is equivalent to: (0.5 * {m + M}) * V^2 = 3490 Hence, their combined velocity V: V^2 = 3490 / (0.5 * 100) = 69.8 V = sqrt(69.8) = 8.354 m/s Retardation due to friction 'a' = 0.6 * g = 5.88 m/s^2 Distance traveled = (V^2) / (2 * a) = (8.354^2) / (2 * 5.88) = 5.934 m Thus, they will travel 5.93 m from the base of the hill

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