The net of a trampoline behaves like an ideal spring. When you, mass of 50kg, st

Question

The net of a trampoline behaves like an ideal spring. When you, mass of 50kg, stand on the trampoline, the net is depressed by .2m. What is the spring constant of the net? When you jump from some height onto the net it is depressed by .4m before coming to a rest momentarily, how much elastic potential energy is stored in the net before it pushes you up? What is your acceleration at this depression? What is the change in your gravitational potential energy when you bounce from max depression to max height?

Explanation / Answer

a) We use the formula F=kx, where F is the force applied by the spring, k is what we are solving for, and x is the amount compressed.

The force the spring is pushing with is equal to the force of gravity pulling down on the person, or mg, so:

mg = kx
(50)(9.8) = k ( .2 )
k = 2450 N/m

b) The energy stored in a spring is U = 1/2 k x2

U = ( 1/2)( 2450 )( .4 )2
U = 196 Joules

c) The acceleration is the sum of all forces, then using F=ma. There is spring force up ( F = kx ) and gravity down ( mg ) so:

kx - mg = ma
(2450)(.4) - ( 50 )(9.8) = 50 a
a = 9.8 ( thats not a coincedence )

d) due to conservation of energy, the total energy stored in the spring will be used to give you height, so the energy you gain from gravitatrional potential energy is EQUAL to the energy given to you by the spring, or 196 Joules.

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