The figure shows an 8.2 kg stone at rest on a spring. The spring is compressed 1

Question

The figure shows an 8.2 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 27 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

Here,

let the spring constant is k

a) for the stone

k * x = m * g

k * 0.11 = 8.2 * 9.8

k = 731 N/m

the spring constant is 731 N/m

b)

compression in spring , x = 0.11 + 0.27 = 0.38 m

elastic potential energy in spring = 0.5 * k * x^2

elastic potential energy in spring = 0.5 * 731 * 0.38^2

elastic potential energy in spring = 52.75 J

c)

as the energy stored in spring initially will change to potential energy of stone earth system

the change in stone earth system = initial elastic potential energy in spring

potential energy of stone earth system = 52.75 J

d)

let the height is h

Using conservation of energy

m *g * h = potential energy of stone earth system

8.2 * 9.8 * h = 52.75

h = 0.656 m = 65.6 cm

the stone rises to maximum height of 65.6 cm

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