The figure shows an 8.3 kg stone at rest on a spring. The spring is compressed 1

Question

The figure shows an 8.3 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 31 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

Given,

m = 8.3 kg ; x = 11 cm

a)We know that for such speing system

weight = Hooks force

mg = kx

k = mg/x = 8.3 x 9.8/0.11 = 739.45 N/m

Hence, k = 739.45 N/m

b)total compression will be:

x = 11 + 31 = 42 cm = 0.42 m

U = 1/2 k x^2

U = 0.5 x 739.45 x 0.42^2 = 65.22 J

Hence, U = 65.22 J

c)from conservation of energy

PE(gravity) = 65.22 J

d)We know that,

PE = m g h

h = PE/mg

h = 65.22/8.3 x 9.8 = 0.802 m

Hence, h = 0.802 m = 80.2 cm

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