(a) How much will a spring that has a force constant of 40.0 N/m be stretched by an object with a mass of 0.500 kg when hung motionless from the spring? (b) Calculate the decrease in gravitational potential energy of the 0.500-kg object when it descends this distance. (c) Part of this gravitational energy goes into the spring. Calculate the energy stored in the spring by this stretch, and compare it with the gravitational potential energy. Explain where the rest of the energy might go.

Explanation:

For equilibrium

mg = kx ( restoring force of spring where k is force constant and x is stretch in it)

x = mg / k

= .5 x 9.8 / 40

= .1225 m

= 12.25 cm

b ) decrease in potential energy

= mgh

= .5 x 9.8 x .1225

= 0.6 J

c ) Energy stored in spring

= 1/2 k x²

= .5 x 40 x ( .1225 )²

= .3 J

It is 50% of gravitational potential energy lost

Rest of the energy might have lost in heating up the spring etc.

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg