Question

In: Physics

Problem 6.37(Mastering Physics Chapter 05: Work and Energy)

A 71 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7 m/s.

Part A 

What is his speed as he lands on the trampoline, 2.2 m below his jump off point?

Part B

If the trampoline behaves like a spring with spring stiffness constant 7.1 * 10^4 N/m , how far does he depress it? Any depression of the trampoline from equilibrium is to be taken as a negative distance.

Solutions

Expert Solution

Part A 

Start by figuring out how high the person will jump. Remember the kinematic equation vf^2 = vi^2 + 2ad. We know that the ending velocity (when he lands on the trampoline) will be zero, so:

vf^2 = vi^2 + 2ad
0^2 = 4.7^2 + 2 * (-9.8) * d
0 = 22.09 – 19.6d
22.09 = 19.6d
d = 1.127 m

So the person will reach a height of 1.127 m above the platform. Since the trampoline is 2.2 m below, they will fall a total of 3.327 m. We can calculate the potential energy and use this to figure out the person’s final speed:

U = mgh
U = 71 * 9.8 * 3.327
U = 2315

Convert to kinetic energy:

KE = 1/2mv^2
2315 = 1/2 * 71 * v^2
2315 = 35.5 * v^2
v^2 = 65.21
v = 8.075 ./s

8.075 m/s

Part B

We found in Part A that the energy when the person hits the trampoline is 2315 J. The potential energy of a spring is 1/2kx^2, so we can solve:

U = 1/2kx^2
2315 = 1/2 * 71000 * x^2
0.065211 = x^2
x = 0.255 m

Remember that mastering physics says this is a negative distance – so the answer is -0.255 m

-0.255 m

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