A 75.0 kg mountain climber is holding his 60.0 kg partner over a cliff when he s

Question

A 75.0 kg mountain climber is holding his 60.0 kg partner over a cliff when he suddenly steps on the ice with a coefficient of kinetic friction of 0.250 at the horizontal top of the cliff, as shown in the accompanying figure. The rope has negligible mass and is held horizontally by the climber. There is no appreciable friction at the icy edge of the cliff. Use energy methods to calculate the speed of the climbers after the lower one has descended 1.80 m starting from rest. What would be the speed of the climbers after they had moved 1.80 m?

(It would really help me if the explanation showed the basic formula and what variables get canceled out throughout the solving process.)

Explanation / Answer

As the lower one moves 1.80 meters down, the upper one slides 1.80 m on the horizontal top of the cliff. As the lower one moves 1.80 meters down, his potential energy his potential energy decreases. ?PE = 60 * 9.8 * 1.8 As the upper one slides 1.80 m on the horizontal top of the cliff, the friction force will decrease the velocity of both climbers. Work done by friction force = µ * m * g * d = 0.3 * 75 * 9.8 * 1.8 The final kinetic energy of both climbers will equal the decrease of PE – the work done by friction force ½ * (75 + 60) * v^2 = (60 * 9.8 * 1.8) – (0.3 * 75 * 9.8 * 1.8) calculate

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