A spring of negligible mass and with spring constant 200.0 N/m is compressed by 0.250 m.Then a steel ball bearing of mass 0.05 kg is put on the spring,and the spring is released.The spring and ball bearing are aligned horizontally and friction is negligible.In addition,assume that the ball bearing will come off the spring exactly as the spring returns to its equilibrium position.The speed of the ball bearing right after the ball bearing loses contact with the spring is

Question

Quizplus · Accepted Answer

First, we need to find the potential energy stored in the compressed spring:
U = (1/2)kx^2
U = (1/2)(200.0 N/m)(0.250 m)^2
U = 6.25 J

When the spring is released, all the potential energy is converted into kinetic energy of the ball bearing:
K = U = 6.25 J

Since the ball bearing is the only object moving, all the kinetic energy is given to it:
K = (1/2)mv^2
6.25 J = (1/2)(0.05 kg)v^2

Solving for v, we get:
v = sqrt(6.25 J / 0.025 kg) = 15.8 m/s

Therefore, the answer is C.

A Spring of Negligible Mass and with Spring Constant 200