A 75 kg climber finds himself dangling over the edge of an ice cliff, as shown i

Question

A 75 kg climber finds himself dangling over the edge of an ice cliff, as shown in the figure below. Fortunately, he's roped to a 880 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.3×10-2. What is his acceleration, and how much time does he have before the rock goes over the edge? Neglect the rope's mass.

What is his acceleration?
How much time does he have before the rock goes over the edge?

http://session.masteringphysics.com/problemAsset/1033731/5/RW-05-44a.jpg

Explanation / Answer

Let T be the tension of the rope => we have : Mg -T =Ma T -µmg =ma => Mg-µmg =(M+m)a => the acceleration : a = (M-µm)g/(M+m) = [ 880 - 75*5.3*10 - 2] 9.8 / [880+75] = 0.92104 m/s2 => the time : L = (1/2)at^2 => t =v(2L/a) = 10.52 s

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