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

Question

A 0.24-kg stone is head 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 5.8 m. 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? 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 J What is the change in gravitational potential energy of the system from release to reaching the bottom of the well? A daredevil on a motorcycle leaves the end of a ramp with a speed of 35.3 m/s as in the figure below. If his speed is 33.2 m/s when he reaches the peak of the Ignore friction and air resistance.

Explanation / Answer

a) GPE = mgh = 0.24 (9.8) (1.2) =2.8224 J

b)GPE = 0.24 ( 9.8) (4.6) =10.8192J

c)change= 13.6416 J

d)1/2 mu^2 = 1/2mv^ 2 +mgh

1/2 ( 35.3)^2 = 1/ 2(33.2)^2 + (9.8) h

h = 7.339 m apprx

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