A Hooke's law spring is compressed a distance d and is used to launch a particle of mass m vertically to a height h above its starting position.Under double the compression,the spring is now used to launch a particle of mass 2m.How high does the second mass rise above its starting position?
A)h
B)2h
C)3h
D)4h

Question

Quizplus · Accepted Answer

According to Hooke's law, the force exerted by a spring is proportional to its displacement. Therefore, if the spring is compressed twice as much, it will exert twice the force.

For the first launch, the force exerted by the spring is F = kx, where k is the spring constant and x is the distance the spring is compressed. This force is equal to the weight of the particle, mg, where g is the acceleration due to gravity. Therefore:

kx = mg

Solving for x:

x = mg/k

The work done by the spring is equal to the potential energy gained by the particle:

W = 1/2 kx^2 = 1/2 mg^2/k

This potential energy is converted into kinetic energy at the moment the particle leaves the spring, and then into potential energy as the particle rises to height h:

1/2 mv^2 = 1/2 mg^2/k

v = sqrt(2g^2/k)

h = v^2/2g = mg/k

For the second launch, the spring is compressed twice as much, so the force exerted by the spring will be twice as much. The particle has twice the mass, so the acceleration due to gravity will be the same as before. Therefore, the particle will rise to a height equal to:

2mg/(2k) = mg/k

This is the same height as before, so the correct answer is B) 2h.

A Hooke's Law Spring Is Compressed a Distance D and Is