How Does a Climber's Mass Relate to Rope Stretch and Fall Distance?

[terrib]

A climber of mass m is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a sping with a spring constant k. He stops to rest, but then accidentally slips and falls freely for a distance h before the rope runs out of slack. The rope then stretches an amount x as it breaks his fall and momentarily brings him to rest. Derive an expression for the climbers mass, m, in terms of k, h, x, and g (the acceleration due to gravity).

When i think about this i automatically think to use the equation:

1/2(k)(y final^2) = (m)(g)(h initial)...
but I do not think that you can use two distances, it wouldn't make sense. Or is the y = to the height?

 

[rl.bhat]

The fall of potential = m*g*(h + x )
The energy stored in the spring = 1/2*k*x^2

 

[ogb p]

Hello,

Thank you for your inquiry. Your initial thought to use the equation 1/2(k)(y final^2) = (m)(g)(h initial) is a good start, but we need to make some adjustments to account for the given information and the situation at hand.

First, let's define our variables:
- m = mass of the climber
- k = spring constant of the nylon rope
- h = distance the climber falls before the rope runs out of slack
- x = amount of rope that stretches as the climber falls
- g = acceleration due to gravity

Now, let's consider the forces acting on the climber. When the climber is at rest, the forces are balanced and there is no acceleration:
- Tension force from the rope (T) = weight force (mg)

When the climber slips and falls, the rope stretches and provides a force to slow the climber's fall:
- Tension force from the rope (T) = weight force (mg) + force from rope stretch (kx)

We can set these two equations equal to each other and solve for m:
T = mg
T = mg + kx
mg = mg + kx
kx = 0
x = 0

This means that the amount of rope that stretches, x, is equal to 0. This makes sense because the climber comes to a stop after falling a distance h and there is no more rope to stretch.

Now, we can plug this value of x into our original equation to solve for m:
1/2(k)(y final^2) = (m)(g)(h initial)
1/2(k)(0^2) = (m)(g)(h)
0 = (m)(g)(h)
m = 0

This means that the mass of the climber is equal to 0, which does not make sense. This is because our initial equation does not take into account the energy lost due to the rope stretching. We need to include this energy in our equation by using the work-energy theorem:

Work done by rope = Change in kinetic energy
- Force from rope stretch (kx)(x) = 1/2(m)(v^2)

We can solve for v, the final velocity of the climber, by using the equation v^2 = u^2 + 2as, where u is the initial velocity