Chapter 08, Problem 019 The figure shows an 8.1 kg stone at rest on a spring. Th

Question

Chapter 08, Problem 019 The figure shows an 8.1 kg stone at rest on a spring. The spring is compressed 8.8 cm by the stone. (a) What is the spring constant? (b) The stone is pushed down an additional 28 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? (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

The force acting on the spring

F = weight of the stone

= mg

= 8.1 * 9.8 = 79.38 N

Spring compression x = 8.8 cm = 0.088 m

a) We know from Hooke's law, |F| = kx

=> k = F/x = 79.38/0.088 = 902 N/m

b) Now, x = 0.088 + 0.28 = 0.368m

Elastic potential energy of the spring:

E = 1/2 * k* x^2

= 1/2 * 902 * 0.368^2

= 61 J

c) Change in PE = change in E

= 1/2 * k * (x1^2 - x2^2)

= 1/2 * 902 * (0.368^2 - 0.088^2

= 57.58 J

d) Change in PE = mgh = 57.58 J

=> h = 57.58/(8.1*9.8) = 0.725m

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