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A 24.0 kg rock, m, is on the edge of a 80.0 m cliff, h, as shown in Figure 11-20. Assume that air resistance is negligible.

(a) What potential energy does the rock possess relative to the base of the cliff?
(b) The rock falls from the cliff. What is its kinetic energy just before it strikes the ground?

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(a) The potential energy of the rock relative to the base of the cliff can be calculated using the formula for gravitational potential energy:

PE = mgh

where m is the mass of the rock (24.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the cliff (80.0 m).

PE = 24.0 kg * 9.8 m/s^2 * 80.0 m = 15,360 J

(b) The kinetic energy of the rock just before it strikes the ground can be calculated using the formula for kinetic energy:

KE = 0.5 * m * v^2

where m is the mass of the rock (24.0 kg) and v is the velocity of the rock just before it strikes the ground. The velocity can be calculated using the equation of motion for a freely falling object:

v = sqrt(2gh)

v = sqrt(2 * 9.8 m/s^2 * 80.0 m) = 44.72 m/s

KE = 0.5 * 24.0 kg * 44.72 m/s^2 = 507.84 J

User Mark Ryan Orosa
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