A 0.23-kg stone is held 1.1 m above the top edge of a water well and then droppe

Question

A 0.23-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 5.6 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 well?

Explanation / Answer

The gravitational potenial energy is given by

P.E=mgH

m-mass of the body,g-acceleration due to gravity,H-height of the body with resect to ground(y=0)

a.)H=1.1m,g=10m/s2,m=0.23kg

P.E=0.23*10*1.1=2.53J

b.)when it reaches bottom of well the height will be H=-5.6m

P.E=-0.23*10*5.6=12.88J

c.) the change in potenial energy is=2.53-(-12.88)=15.41J

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