A 1 kg mass is attached to a vertical spring, stretching it by 10 cm from its equilibrium position. With the spring secured in the same stretched position, the 1 kg mass is removed and replaced with a 0.50 kg mass. If the spring is then released, how high will the 0.50 kg mass rise above the point where it was released?
A) 0.10 m
B) 0.24 m
C) 0.44 m
D) 0.80 m
E) 1.0 m

Question

Quizplus · Accepted Answer

The potential energy stored in the spring is equal to the work done in stretching it. When the 1 kg mass is attached, the work done is:$$W = Fd = mgd = (1 kg)(9.8 m/s^2)(0.10 m) = 0.98 J$$When the 0.50 kg mass is attached, the work done is:$$W = Fd = mgd = (0.50 kg)(9.8 m/s^2)(d) = 4.9d J$$Since the spring is secured in the same stretched position, the work done in both cases is the same. Therefore, we have:$$0.98 J = 4.9d J$$Solving for d, we get:$$d = 0.20 m$$Therefore, the 0.50 kg mass will rise to a height of 0.20 m above the point where it was released.

A 1 Kg Mass Is Attached to a Vertical Spring