Rock Climber Sliding Down a Smooth Slab 9.6 m Hk = 0.72 %3D 56° Consider the rock climber climbing on a smooth slab, as shown above: the coefficient of kinetic friction between the climber's rock shoes and the rock is 0.72 the belayer cannot see the climber who is out of sight beyond the edge of the slab the belayer cannot hear the climber because the wind is blowing very hard The climber loses his footing and slides down the slab from rest. How long after the climber loses his footing does the belayer realize she needs to catch the climber’s fall? [answer: about 2.1 s] FNET = ma W = mg g= 9.81 m/s2 Ax = vot + ½ at2
Rock Climber Sliding Down a Smooth Slab 9.6 m Hk = 0.72 %3D 56° Consider the rock climber climbing on a smooth slab, as shown above: the coefficient of kinetic friction between the climber's rock shoes and the rock is 0.72 the belayer cannot see the climber who is out of sight beyond the edge of the slab the belayer cannot hear the climber because the wind is blowing very hard The climber loses his footing and slides down the slab from rest. How long after the climber loses his footing does the belayer realize she needs to catch the climber’s fall? [answer: about 2.1 s] FNET = ma W = mg g= 9.81 m/s2 Ax = vot + ½ at2
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Rock Climber Sliding Down a Smooth Slab**
*Diagram Description:*
The diagram illustrates a scenario where a rock climber is situated on a smooth slab inclined at an angle of 56° from the horizontal. The length of the slab is 9.6 meters. The coefficient of kinetic friction (\( \mu_k \)) between the climber's rock shoes and the slab is 0.72. The diagram also depicts a belayer at the bottom, unable to see the climber due to the position and the edge of the slab.
*Problem Statement:*
Consider the rock climber climbing on a smooth slab, as shown above:
- The coefficient of kinetic friction between the climber’s rock shoes and the rock is 0.72.
- The belayer cannot see the climber who is out of sight beyond the edge of the slab.
- The belayer cannot hear the climber because the wind is blowing very hard.
The climber loses his footing and slides down the slab **from rest**. How long after the climber loses his footing does the belayer realize she needs to catch the climber’s fall? [answer: about 2.1 s]
*Equations and Variables:*
- \( F_{\text{net}} = ma \)
- \( f_k = \mu_k N \)
- \( W = mg \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( \Delta x = v_0 t + \frac{1}{2} at^2 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe3e68242-5a91-4b54-b664-0ff90b744d38%2Fe1759305-d513-444c-a983-7e93b3993372%2Fcrn4x5a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Rock Climber Sliding Down a Smooth Slab**
*Diagram Description:*
The diagram illustrates a scenario where a rock climber is situated on a smooth slab inclined at an angle of 56° from the horizontal. The length of the slab is 9.6 meters. The coefficient of kinetic friction (\( \mu_k \)) between the climber's rock shoes and the slab is 0.72. The diagram also depicts a belayer at the bottom, unable to see the climber due to the position and the edge of the slab.
*Problem Statement:*
Consider the rock climber climbing on a smooth slab, as shown above:
- The coefficient of kinetic friction between the climber’s rock shoes and the rock is 0.72.
- The belayer cannot see the climber who is out of sight beyond the edge of the slab.
- The belayer cannot hear the climber because the wind is blowing very hard.
The climber loses his footing and slides down the slab **from rest**. How long after the climber loses his footing does the belayer realize she needs to catch the climber’s fall? [answer: about 2.1 s]
*Equations and Variables:*
- \( F_{\text{net}} = ma \)
- \( f_k = \mu_k N \)
- \( W = mg \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( \Delta x = v_0 t + \frac{1}{2} at^2 \)
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