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## Homework Statement

A 75 kg climber finds himself dangling over the edge of an ice cliff. Fortunately, he's roped to a 920 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? Neglect the rope's mass.

## Homework Equations

Tension(rock)=tension(climber)=T

acceleration(rock)=acceleration(climber)=a

Climber(y):T-m(c)*g=m(c)*a

Rock(x):T+f(k)=m(r)*a

Rock(y):N-m(r)*g=0

## The Attempt at a Solution

T=m(c)*a+m(c)*g

f(k)=mu(k)*N

N=m(r)*g

m(c)*a+m(c)*g+mu(k)*m(r)*g=m(r)*a

a=[m(c)*g+mu(k)*m(r)*g]/[m(r)-m(c)]

a=1.44 m/s^2

The computer is telling me that I have the wrong answer. Do I need to take the 51 m into account somehow? Any help would be greatly appreciated!