Rock climbers seek safety by placing \"protection\" in the rock. These are fixed

Question

Rock climbers seek safety by placing "protection" in the rock. These are fixed pieces of equipment called "cams" or "friends" that their rope passes through. A climber's partner holds the other end of the rope, so if she falls, the protection catches and stops her fall. In some situations, a climber falls far enough that the force exerted on the protection pulls it out of the rock face. The climber then falls farther, and exerts even more force on the next piece of protection, and pulls it out too. This terrifying phenomenon is called "zippering a pitch."

Imagine a climber has climbed 5.00 m directly above her uppermost piece of protection, with no slack (excess length) in the rope, and the next piece of protection is 6.00 m directly below the uppermost piece of protection. She falls with an acceleration of 9.80 m/s2 downward. The first piece of protection cuts her velocity in half, but then it fails and she falls farther. What is her speed just when the rope pulls on the second piece of protection? (#2.A.4)

Please answer in m/s

Rock climbers seek safety by placing "protection" in the rock. These are fixed pieces of equipment called "cams" or "friends" that their rope passes through. A climber's partner holds the other end of the rope, so if she falls, the protection catches and stops her fall. In some situations, a climber falls far enough that the force exerted on the protection pulls it out of the rock face. The climber then falls farther, and exerts even more force on the next piece of protection, and pulls it out too. This terrifying phenomenon is called "zippering a pitch."

Imagine a climber has climbed 5.00 m directly above her uppermost piece of protection, with no slack (excess length) in the rope, and the next piece of protection is 6.00 m directly below the uppermost piece of protection. She falls with an acceleration of 9.80 m/s2 downward. The first piece of protection cuts her velocity in half, but then it fails and she falls farther. What is her speed just when the rope pulls on the second piece of protection? (#2.A.4)

Please answer in m/s

Explanation / Answer

Conserving energy

When she falls to first piece of protection, she has covered 5 + 5 = 10m

So conserving energy

mgh = 0.5mv^2

v = (2gh)^0.5 = 14 m/s

Since first protection cuts her speed by half = 7m/s

When second protection holds her, she will be 1+6 = 7m further down

So Conserving energy again

0.5m(v/2)^2 + mgh = 0.5mv'^2

0.5*7^2 + 9.8*7 = 0.5*v'^2

v' = 13.645 m/s

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.