Chapter 4, Quiz 2 Sep 17 2018 A rock climber, weighing 80kg, falls and is caught

Question

Chapter 4, Quiz 2 Sep 17 2018 A rock climber, weighing 80kg, falls and is caught by his rope. The last point the rope was anchored to the rock is 1 m below the point where the rope is attached to the climber at the time he slips. Hence the climber falls 2m straight down before his rope starts stretching How fast is the climber moving at this point (assume acceleration due to gravity, g = 9.8m/s*)? 3.1 m/s 4.4 m/s 6.3 m/s 9.8 m/s If his rope can stretch by at most 20%, what is the average acceleration (absolute value) of the climber while the rope is stopping his fall? 0.2 × g 10×g What will be the average tension in the rope while it is saving the climber's life? 1.6 kN (160kg x g) 5 x g 8.6 kN (880 kg × g) 7.8 kN (800 kg × g) 078 kN (80kg × g) If instead the rope could only stretch 5%, what would the average tension be? 32 kN (3.3 tons x g) 8.6 kN (0.88 tons x g) 1.6 kN (160kg x g) 0.78 kN (80kg x g)

Explanation / Answer

2. When the rope stretches by 20 percent, the distance of fall of the climber from the point when the rope is in stretched, that is, the last point in question (1) to the point when the rope stops the climber is

= 20 percent of 1m = (20/100 ×1) m = 0.2 m

Now using work - energy theorem, we have,

Change in kinetic energy = work done

1/2 mv^2 = mah (since W = F×s)

a = v^2/2h

Here, v = 6.3 m/s and h = 0.2 m

So, average acceleration,a = (6.3)^2/(2×0.2) = 98 m/s^2

   = 10×9.8 m/s^2 = 10g

3. Tension in the rope is equal to the the net downward force on the rope by the falling climber, so,

Tension, T = Force = ma

Here, m = 88 kg, a = 10g

Thus, T = 88×10g = 880g

= 880×9.8 = 8624 = 8.6×10^3 N = 8.6 km

Since, 1 kN = 10^3 N

4. When the rope stretched by 5 percent, as explained in question (2), height,h = (5/100 × 1) m = 0.05 m

So, acceleration, a = v^2/2h = (6.3)^2/2×0.05

= 392 m/s^2

So, tension, T = ma = 88×392 N = 3.2 kN

  

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