A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position.The spring constant of the spring is 1.00 N/cm.The mass is pulled downward 2.00 cm and released.What is the speed of the mass when it is 1.00 cm above the point from which it was released?
A) 0.0443 m/s
B) 0.0744 m/s
C) 0.0201 m/s
D) 0.0866 m/s
E) The mass will not reach the height specified.

Question

Quizplus · Accepted Answer

First, we need to find the potential energy stored in the spring when the mass is pulled down by 2.00 cm:

U = 1/2 k x^2 = 1/2 (1.00 N/cm)(0.02 m)^2 = 0.0002 J

This is also the amount of kinetic energy that the mass will have at the equilibrium position (when it is released).

When the mass is 1.00 cm above the equilibrium position, it will have lost some of its potential and kinetic energy to gravitational potential energy.

At this point, we can use conservation of energy to find the velocity of the mass:

U(initial) = U(final) + K(final)

0.0002 J = mgh + 1/2 mv^2

Using h = 0.02 m - 0.01 m = 0.01 m (the distance between the equilibrium position and the final height) and solving for v, we get:

v = sqrt[(2/m)(0.0002 J - mgh)] = 0.0866 m/s

Therefore, the answer is D.

A 400-Kg Mass Is Attached to a Very Light Ideal Spring