Question

Patrick is repairing the roof on his house. He uses a hammer having a mass of 1 kilogram. While working at the apex of the roof, the hammer slips and falls to the ground at a velocity of 12 meters/second. Find the potential energy of the hammer just before slipping.

Answer 1

The potential energy is 72 J

Step-by-step explanation:

PEA = mghA = (1)(9.8)(hA) = 9.8 hA

KEA = mv2 = (1)(12)2 = 72 J

PEG = mghG = (1)(9.8)(hG) = 9.8(0) = 0

By the law of conservation of energy,

Total energy of the hammer = TE = PEA + KEA = PEG + KEG

PEA + 0 = 0 + 72

PEA = 72 J

So the potential energy of the hammer just before slipping is 72 J

Answer 2

72 joules

Step-by-step explanation:

The potential energy of that hammer is a function of its displacement against gravity. Considering that it fell with a velocity of 12 m/s, it was its displacement against gravity that gave it this velocity. It will continue to move until its displacement to gravity is zero.

since the body is in motion; it has converted its potential energy (mgh, m is mass, g is acceleration due to gravity, and h is the height) to kinetic energy (energy due to motion, 1/2mv^2; m = mass, v = velocity or speed)

therefore the potential energy is equal to kinetic energy

mgh = 1/2mv^2 = 1/2 *1kg* 12*12 = 72 joules.