Part a
two rock climbers, jim and karen, use safety ropes of simi-lar length. karen's rope is more elastic, called a dynamic rope by climbers. jim has a static rope, not recommended for safety purposes in pro climbing. karen falls freely about 2.0 m and then the rope stops her over a distance of 1.0 m. (figure 1)
part a
estimate how large a force (assume constant) she will feel from the rope. express the result in multiples of her weight.
express your answer using two significant figures.
f =

You need to find the acceleration once the rope starts acting.

For that, first you need the velocity, V, when Karen ends the 2.0 m free fall

V^2 = Vo^2 + 2gd = 0^2 + 2* 9.81m/s^2 * 2.0 m = 39.24 m^2 / s^2 => V = 6.26 m/s

From that point, the rope starts acting with a net force that produces a constant acceleration, modeled as per these values and equation:

Vo = 6.26m/s
Vf = 0

Vf^2 = Vo^2 + 2ad => a = [Vf^2 - Vo^2] / 2d = [0^2 - (6.26 m/s)2 ] /2(1m) = -19.62 m/s^2

That acceleration is due to the difference of the force applied by the rope - the weight of Karen:

Weight of Karen: mg

Weight of karen - Force of the rope = Net force = ma

mg - Force of the rope = ma

Force of the rope = mg - ma = m(g -a) = m [9.81m/s^2 - (-19.62m/s^2) ]=  29.43m

To express it as a multiple of her weight, divide it by her weight (mg = 9.81m) =>

29.43m / 9.81m = 3.0

3.0 times.