Question

20. A 15.9kg rock is dropped from a height of 113m. Calculate its potential energy. What is the rock'skinetic energy right before it hits the ground? What is the rock's velocity right before it hits the ground?

Answer 1

Kinetic energy = 1/2 * m * v^2

Potential energy = m*g*h

m = 15.9 kg

h= 113 m

Potential energy = 15.9 kg * 9.8 N * 113 m = 17,607.66 J

Kinetic energy right before it hits the ground

PE = KE

mgh = 1/2 m v^2

Masses cancel out

gh = 1/2v^2

9.8 N * 113m = 1/2 v^2

Solve for v

1,107.4 Nm = 1/2 v^2

1,107.4 Nm / (1/2) = v^2

2,214.8 Nm = v^2

√2,214 = v

47.06 m/s = V

Kinetic energy : 1/2 * m * v^2 = 1/2 * 15.9 * 47.06^2 = 17,606.41 J