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 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? Number Units Number Units Number Units Number Units

Explanation / Answer

given that

m = 8.3 kg

x = 11 cm = 0.11 m

we know that

F = k*x ....eq1

weight of stone F = m*g .....eq2

from eq 1&2

k*x = m*g

k = m*g/x = 8.3*9.8/0.11

k = 739.45 N/m

(b)

total compression x' = 11+33 = 44 cm = 0.44 m

elastic potetia energy PEe = k*x'^2 / 2

PEe = 739.45*(0.44)^2 / 2 = 71.57 J

(c)

stone is released and the stored elastic energy is completely converted to gravitational potential energy.

so, change in gravitational potential energy is

dPEg = 71.57 J

(d)

dPEg = m*g(hmax-0)

m*g*hmax = 71.57

hmax = 71.57 / 8.3*9.8

hmax = 0.88 m

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