Question

A 120-g mass is attached to the bottom end of an unstressed spring. The spring is hanging vertically and has a spring constant of 40.0 N/m. The mass is dropped. How far does it drop before coming to rest momentarily?

Answer 1

The mass drops 0 meters before coming to rest momentarily.

Step-by-step explanation:

To find the distance the mass drops before coming to rest momentarily, we can use the equation for gravitational potential energy:

PE = mgh

Where PE is the gravitational potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the gravitational potential energy is equal to the energy stored in the spring. We can set these two equal:

PE = (1/2)kx^2

Where k is the spring constant and x is the displacement of the mass from its equilibrium position. Rearranging the equation, we get:

x = sqrt((2PE)/k)

Substituting the known values, we have:

x = sqrt((2(0.120 kg)(9.8 m/s^2)(h))/(40.0 N/m))

x = sqrt(0.588h)

Now we need to find the value of h where the mass comes to rest momentarily. This occurs when the gravitational potential energy is equal to zero, since the mass has dropped all of its potential energy. Thus, we can set:

0.588h = 0

And solving for h:

h = 0

Therefore, the mass drops 0 meters before coming to rest momentarily.