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A 70-kg person, starting from rest, jumps vertically downward onto a trampoline

ID: 1427843 • Letter: A

Question

A 70-kg person, starting from rest, jumps vertically downward onto a trampoline 3-m below his starting point. Assume that the trampoline acts like a spring, and that it stretches 0.5 m in bringing the person to rest. Neglect all resistive forces (i.e., air resistance and friction).
(a) What is the initial gravitational PE of the person relative to when he is at rest at the bottom of the trampoline?

(b) What is the spring constant of the trampoline? (Hint: Use conservation of energy).

(c) What will be the person’s speed when he just leaves the net after rebounding?

Explanation / Answer

a) the person travels a total of 3.5m downward before coming to absolute rest which we take to be the ground state of potential. So potential enrgy relative to ground = mgh J = 2401 J

b) the person fell freely for 3 m, so velocity just before touching net = sqrt(2*g*h) m/s = 7.66 m/s

so KE = 0.5*m*v2 J = 2058 J

This is completely absorbed by the spring, whose spring constant is say k, so

k*0.52 = 2058 or k = 8232 N/m

c) the person's speed when leaving the net = vel when he was incoming = 7.66m/s

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