A 1.6 kg rock initially at rest loses 404 J of potential energy while falling to the ground. Assume that air resistance is negligible.
Calculate the kinetic energy that the rock gains while falling.
What is the rock's speed just before it strikes the ground?
m/s

Explanation:

The point about the question is that if PE is lost, then KE must be gained.

So the answer to the first part is 404 Joules.

Unless the total kinetic energy gained is 404 joules, the problem is impossible to calculate the second part. We'll assume that the 404 is the total amount of energy gained by the object.

KE = 404

PE = 0

KE = 1/2 m v^2

KE = 404

m = 1.6

404 = 1/2 1.6 kg * v^2       Multiply 1.6 and 1/2 which is 0.8

404 = 0.8 kg * v^2            Divide both sides by 0.8

404/0.8 = v^2

505 = v^2                          Take the square root of both sides.

v = 22.47 m/s

Just before hiting the ground the speed is 22.47 m/s

free point tnx…:)