Question

# A mountain climber of mass m=75.0 kg slips during a climb and begins to fall freely....

A mountain climber of mass m=75.0 kg slips during a climb and begins to fall freely. She forgot her rope and had to use an ideal spring. The spring is hanging vertically with an unstretched equilibrium position at the origin, x=0. On her descent, she grabs the end of the spring which has a spring constant of k= 981 N/m. Ignore friction, air resistance, and any other dissipative effects.

a) The climbers velocity when she grabbed the spring was 20.0 m/s. How far down does she travel before coming to rest? Justify your answer using the conservation of energy equation: Ei=Ef=1/2mvo^2 + mgxo = 1/2 kxo^2.

b) If she continues holding on, what is the maximum height she will go?

c) If she continues holding on, at what height (i.e. position) will her velocity be maximum?

d) If she continues holding on, how much time elapses traveling from her highest point to her lowest point?

e) What is the initial height from which she fell? Is there a way she can reach that point by letting go of the spring? If so, at what point int he oscillation should she let go?

Please show all work with solutions. Thank you!

Here we apply energy conservation and simple harmonic motion concept.

#### Earn Coins

Coins can be redeemed for fabulous gifts.