Question

Suppose you have a spring with a force constant of 42.5 N/m.

How much is the spring stretched, in meters, by an object with a mass of 0.49 kg that is hanging from the spring at rest?

Calculate the change in gravitational potential energy, in joules, of the 0.49-kg object when it descends this distance.

Calculate the energy, in joules, stored in the spring when it is stretched this amount.

Apparently, not all of the gravitational potential energy went into the potential energy of the spring. What happened to the rest of the potential energy?

Answer 1

k = force constant of the spring = 42.5 N/m

m = mass of the object hanged = 0.49 kg

x = stretch in the spring caused ?

using equilibrium of force along the vertical direction, we get

k x = mg

(42.5) x = (0.49)(9.8)

x = 0.113 m

U = change in gravitational potential energy of the object

change in gravitational potential energy of the object is given as

U = - mgx

U = - (0.49)(9.8)(0.113)

U = - 0.543 J

Spring potential energy of the spring is given as

U_s = (0.5) k x²

U_s = (0.5) (42.5) (0.113)²

U_s = 0.271 J

rest of the energy is lost.