- #1

josevie

- 5

- 3

- Homework Statement
- An 8.00kg stone is at rest on a spring. The spring is compressed 10.0cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 30.0cm and released. What is the elastic potential energy of the compressed spring just before that release? (c) What is the change in the gravitational potential energy of the stoneEarth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?

- Relevant Equations
- F=m.a =-k.x = m.g

Us = (1/2).k.x²

Ug = mgh

It is to my understanding that if the spring was compressed 10cm, it is due to the Work of the Weight Force of the stone. So:

Work done on the spring by the stone = m.g.x = 7.84 J

The work done on the spring will be stored as potential energy of the spring, so:

Us = W

Us = (1/2).k.x²

k = 2.Us/x² which gives me k = 1568 N/m, which is double the right answer of 784 N/m

I know that if I were to use Newton's second law and F(spring) = m.g, I would find the right answer. I cant, however, understand why the above method is wrong. Someone please help lol

Work done on the spring by the stone = m.g.x = 7.84 J

The work done on the spring will be stored as potential energy of the spring, so:

Us = W

Us = (1/2).k.x²

k = 2.Us/x² which gives me k = 1568 N/m, which is double the right answer of 784 N/m

I know that if I were to use Newton's second law and F(spring) = m.g, I would find the right answer. I cant, however, understand why the above method is wrong. Someone please help lol