A Hooke's law spring is mounted horizontally over a frictionless surface.The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface.What compression of the spring would result in the mass attaining 4 times the speed received in the situation described here?
A)2.00d
B)0.50d
C)4.00d
D)16.00d

Question

Quizplus · Accepted Answer

According to conservation of energy, the initial potential energy stored in the spring equals the final kinetic energy of the mass.

Let's assume the initial speed of the mass is v. Then, we have:

(1/2)k(d^2) = (1/2)m(v^2)

where k is the spring constant.

If we want the final speed of the mass to be 4v, we can use the same equation:

(1/2)kx^2 = (1/2)m(16v^2)

where x is the new compression of the spring.

Dividing the second equation by the first, we get:

x^2/d^2 = 16

x = 4d

Therefore, the compression of the spring required to attain 4 times the initial speed is 4d. So, the answer is (C).

A Hooke's Law Spring Is Mounted Horizontally Over a Frictionless