Conservation of energy

Question

Camera has a 800 KG Boulder, a top of a 40m hill. If she pushes the Boulder off the hill with an initial velocity of 1 M/S, how fast will it be rolling when it reaches the bottom of the hill?

Answer 1

Initial kinetic energy (KE) and potential energy (PE)

KE₀ = ½m(v₀)² = ½m(800)(1)² = 400 J

PE₀ = mgh = (800)(9.8)(40) = 313,600 J

Toral energy = 313,600 + 400 = 314,000 J

Assuming no energy is lost to nonconservative forces (like friction), at the bottom you will have 314,000 J of kinetic energy.

So ½mv² = 314,000 which means

v² = 2(314,000 J)/(800 kg) = 785 m²/s²

and

v = √(785) ≈ 28 m/s