The figure shows an 8.3 kg stone at rest on a spring. The spring is compressed 9

Question

The figure shows an 8.3 kg stone at rest on a spring. The spring is compressed 9.8 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

given,

mass of stone = 8.3 kg

spring is compressed by = 9.8 cm

so,

mg = kx

8.3 * 9.8 = k * 0.098

spring constant k = 830 N/m

stone is pushed down additionally 33 cm then,

x = 9.8 + 33

x = 42.8 cm

elastic potential energy = 0.5 * k * x^2

elastic potential energy = 0.5 * 830 * 0.428^2

elastic potential energy = 76.02 J

suppose when spring is compressed to 42.8 cm the potential energy of the stone is 0 J so when the spring is released the energy lost by the spring is gained by the stone and final potential energy of the stone will become 76.02 J

mgh = 76.02

8.3 * 9.8 * h = 76.02

h = 0.935 m

maximum height, measured from the release point = 0.935 m

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