Final answer:
The 5.0 kg bowling ball will move 2.0 m below the equilibrium surface of the trampoline when dropped from a height of 3.0 m onto an ideal spring with a spring constant of 500 N/m.
Step-by-step explanation:
The question asks how far down a 5.0 kg bowling ball moves below the equilibrium surface of the trampoline after being dropped from a height of 3.0m, where the trampoline is modeled as an ideal spring with a spring constant of 500 N/m. To solve this, we must equate the gravitational potential energy of the bowling ball at the height it is dropped from to the elastic potential energy of the spring when the ball has come to a stop momentarily.
The formula for gravitational potential energy (PEg) is PEg = mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height from which the ball is dropped. The formula for the elastic potential energy (PEs) of a spring is PEs = 1/2 kx2, where k is the spring constant and x is the displacement from the equilibrium position. Setting these two equal gives mgh = 1/2 kx2.
Plugging in the values, we get (5.0 kg)(9.81 m/s2)(3.0 m) = 1/2 (500 N/m)x2. Solving for x, we find that x = 2.0 m, which corresponds to choice (b) 2.0 m as the distance the ball moves below the equilibrium surface of the trampoline.