A 0.24-kg stone is held 1.2 m above the top edge of a water well and then droppe

Question

A 0.24-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 5.5 m (a) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system before the stone is released? (b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well? (c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the wel? Need Help? Read N

Explanation / Answer

the gravittaional potential energy is given by'

PE = mgh

where,

m=mass of the object = 0.24 kg

g= accleration due to gravity = 9.81 m/s2

and h=height

a) for the case before the stone is released, h=1.2m

therefore gravitational potential energy is

PE = mgh = (0.24)(9.81)(1.2) = 2.822 Joules

b) for this case when the stone is released h = 1.2+5.5 = 6.7m

thus potential energy is

PE = (0.24)(9.81)(-6.7) = -15.77 Joules

it is negative because it is acting downwards

c) change in the gravitational potential energy is

-15.77-2.82 = -18.59 Joules

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