Chapter 08, Problem 019 The figure shows an 8.0 kg stone at rest on a spring. Th

Question

Chapter 08, Problem 019

The figure shows an 8.0 kg stone at rest on a spring. The spring is compressed 11 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 33 cm 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 stone–Earth system when the stone moves from the release point to its maximum height? (d) What is that maximum height, measured from the release point?

Explanation / Answer

A) F = kx , for a spring constant k and displacement from the equilibrium position x and restoring force F.

Equate F to the weight of the stone: F = mg = 8 x 9.8 = 78.4 which is equal to kx or k x 0.11:
=> 0.11k = 78.4
=> k = 712.727 N/m
as you say.

B) Stored elastic potential energy is given by U_e = 1/2 kx^2 (integrated force x displacement).

U_e = 1/2 x 712.727 x (0.44)^2 -(0.11^2)= 64.68 J
This one's right too.

C) The stone is released and the stored elastic energy is completely converted to gravitational potential energy so the change in gravitational potential energy (DU_g) is:

DU_g = U_e = 64.68 J.

(The stone's kinetic energy at the maximum height is 0.)

D) Since change in gravitational potential energy is given by DU_g = mg(Dx) where Dx is the change in height we have:

mg(Dx) = 64.68 j
=> Dx = 64.68/(8 x 9.8) = 0.827 m is the height from the release point

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