A rock falls from a vertical cliff that is 4.0 m tall.At what speed will the gravitational potential energy of the rock-Earth system (relative to the base of the cliff)be equal to its kinetic energy? Assume that air has no effect on the motion of the rock.
A)3.1 m/s
B)4.4 m/s
C)6.3 m/s
D)8.9 m/s
E)13 m/s

Question

Quizplus · Accepted Answer

The gravitational potential energy (PE) of the rock at the top of the cliff is equal to its kinetic energy (KE) when it has fallen a distance where these two energies are equal. This can be found using the conservation of energy principle. The potential energy at the top is given by PE = mgh, where m is the mass of the rock, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the cliff (4.0 m). The kinetic energy at any point is given by KE = 1/2 mv^2, where v is the velocity of the rock. Setting PE equal to KE gives:mgh = 1/2 mv^2Solving for v (and noting that m cancels out), we get:v = sqrt(2gh) = sqrt(2*9.8*4.0) = sqrt(78.4) ≈ 8.9 m/s.However, the question asks for the speed when the gravitational potential energy is equal to its kinetic energy, which occurs halfway through the fall, not at the bottom. At this point, only half of the potential energy has been converted to kinetic energy. To correct my calculation mistake:For the energies to be equal, we only need to consider the point where the rock has fallen half the distance, because energy conversion is linear in this context. Therefore, we should calculate the speed at the halfway point using the correct approach for energy equality. The correct approach involves using the principle of energy conservation from the start, but with the understanding that the potential energy lost equals the kinetic energy gained at any point in the fall, not just at the halfway or the end.Given the mistake in the initial explanation, let's correctly solve for the speed when the potential energy equals the kinetic energy without assuming it's at the halfway point, as this assumption was incorrect:Using the formula for potential energy (PE = mgh) and kinetic energy (KE = 1/2 mv^2) and setting them equal to each other for the full height (since we are looking for the point where they are equal, not assuming it's at the halfway):mgh = 1/2 mv^2Solving for v gives:v = sqrt(2gh) = sqrt(2*9.8*4.0) = sqrt(78.4) ≈ 8.9 m/s.This correction maintains the conclusion that the speed when the gravitational potential energy equals the kinetic energy, considering the entire height, is approximately 8.9 m/s, which matches option D, not C as initially stated. The mistake was in the interpretation of the question and the explanation, not in the calculation per se. The correct answer, based on the corrected interpretation that considers the energy equality over the entire height, should indeed be 8.9 m/s, aligning with option D.

A Rock Falls from a Vertical Cliff That Is 4