Kevin S. Huang

2013: Problems 23-24

Topic: Energy
Concepts: Conservation of energy, Forces in mechanics

Solution:

By conservation of energy, the gravitational potential energy of the man is converted into spring potential energy at the bottom of his fall:

mgH=\frac{1}{2}kh^2

This is also when the tension in the spring is maximized so

T=kh=4mg

h=\frac{4mg}{k}

Substituting this into the first equation,

mgH=\frac{1}{2}k\left(\frac{4mg}{k}\right)^2=\frac{8m^2g^2}{k}

Thus,

k=\frac{8mg}{H}

so the answer is E.

Topic: Dynamics
Concepts: Forces in mechanics

Solution:

From the previous problem, we found the spring constant is

k=\frac{8mg}{H}

We are given the maximum tension in the cord is

T=kh=4mg

h=\frac{4mg}{k}=\frac{4mg}{8mg/H}=\frac{H}{2}

Thus, the answer is A.