Solve Kinetic Friction Homework: Acceleration & Time Before Rock Goes Over Edge

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SUMMARY

The discussion centers on a physics homework problem involving a 75 kg climber and a 980 kg rock on an ice cliff, with a kinetic friction coefficient of 0.055. The climber's acceleration was incorrectly calculated as -2.76 m/s² due to a misunderstanding of the forces acting on the rock. The correct approach requires applying the equation of motion \(\sum F = ma\) for the rock to determine the accurate tension and subsequent acceleration. The climber's acceleration and the time before the rock falls over the edge must be recalculated using the correct equations.

PREREQUISITES
  • Understanding of Newton's second law of motion (F=ma)
  • Knowledge of kinetic friction and its coefficient
  • Basic algebra for solving equations
  • Familiarity with free-body diagrams
NEXT STEPS
  • Review the concept of free-body diagrams in physics
  • Learn how to apply \(\sum F = ma\) in multi-body systems
  • Study the effects of kinetic friction on motion
  • Practice similar problems involving tension and acceleration in physics
USEFUL FOR

Students studying physics, particularly those tackling problems involving forces, tension, and friction in multi-body systems.

orange03
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Homework Statement



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 980 kg rock located 51 m from the edge of the cliff. Assume that the coefficient of kinetic friction between rock and ice is 5.5×10−2. 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?




Homework Equations


F=ma


The Attempt at a Solution



equation of the climber:
Ft-mg=may

equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N. I plugged this back into the climber's equation and got -2.76 m/s^2 but this answer is wrong. Help please!
 
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orange03 said:
equation of the rock:
Ft=Fr

I found the tension of the ropes by this equation: Ft=ukN=(.055)(980)(9.8) = 528.22N.

The equation of the rock is wrong
The tension is not the same as friction because the rock moves with acceleration.
Use \sum F = ma to get the equation for the rock
 

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