The figure shows an 8.00 kg stone resting on a spring. The spring is compressed

Question

The figure shows an 8.00 kg stone resting on a spring. The spring is compressed 8.0 cm by the stone.
http://www.webassign.net/hrw/08_35.gif
(a) What is the spring constant?
N/cm

(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
J

(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?
J

(d) What is that maximum height, measured from the release point?
m

Explanation / Answer

The mass of stone, m = 8 kg The compression in the spring, x = 8 cm = 0.08 m a) We have a formula for the spring force, F = kx From the above equation, k = F/x = mg/x = 980 N/m = 9.8 N/cm b) The total compression in the spring, x = 30 + 8 = 38 cm = 0.38 m The spring constant, k = 980 N/m The elastic potnetial energy of spring, U = (1/2)kx^2                          U = 0.5 * 980 * 0.38 * 0.38 = 70.76 J c) Initially the potential energy of spring is equal to the kinetic energy of the stone    (1/2)mv^2 = (1/2)kx^2 From the above, v^2 = kx^2/m = 17.689 Then the velocity, v = 4.2 m/s So the maximum height reached by the stone, h = v^2/2g = 0.9025 m Now the change in gravitational potential energy = mgh = 8 * 9.8 * 0.9025                                                   = 70.756 J

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