Question

A 5.0 kg bowling ball is dropped from rest 3.0 m above a trampoline. The trampoline can be modeled as an ideal spring with a spring constant of 5.0E+2 N/m. How far down does the bowling ball move below the equilibrium surface 0 the trampoline? Ignore air resistance, etc.

Answer 1

The bowling ball moves 0.49 meters below the equilibrium surface of the trampoline.

When the 5.0 kg bowling ball is dropped from a height of 3.0 meters above the trampoline, it accelerates due to gravity and gains kinetic energy. As the ball makes contact with the trampoline, it compresses the spring in the trampoline. The potential energy gained by the ball during its fall is transferred into the potential energy stored in the compressed spring.

The potential energy (PE) of the bowling ball at its highest point is given by PE=mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. The potential energy is then converted to spring potential energy (PE spring) given by PE spring = 1/2 kx^2, where k is the spring constant and xis the displacement from the equilibrium position.

Setting these two potential energies equal to each other and solving for

x, we find x=
`√(2mgh/k)`. Substituting the given values into this formula results in x≈0.49 meters. Therefore, the bowling ball moves approximately 0.49 meters below the equilibrium surface of the trampoline. This calculation assumes no energy losses to factors such as air resistance or friction in the trampoline.