Question

g Case 1, a mass M M hangs from a vertical spring having spring constant k , k, and is at rest at its equilibrium height. In Case 2, the same mass has been lifted a distance D D vertically upward. If the potential energy in Case 1 is defined to be zero, what is the potential energy in Case 2

Answer 1

Answer:

$U = MgD + \frac{1}{2}kD^2$

Explanation:

In the first case, the mass is at equilibrium. Therefore, the net force on the mass is equal to zero.

In the second case, the spring is compressed by an amount of D, which is equal to an elastic potential energy of

$U_e = \frac{1}{2}kD^2$

At this height, the gravitational potential energy of the mass is equal to

$U_g = MgD$

Then, the total potential energy at height D is

$U = MgD + \frac{1}{2}kD^2$