force and potential energy

Question

A roller coaster cart of mass m = 221 kg starts stationary at point A, where h1 = 28.9 m and a while later is at B, where h2 = 13.1 m. The acceleration of gravity is 9.8 m/s 2 .What is the potential energy of the cart relative to the ground at A?

a) Answer in units of J.

What is the speed of the cart at B, ignoring the effect of friction?

b) Answer in units of m/s

Answer 1

KE₁ + PE₁ = KE₂ + PE_{2 ( m = 221 kg, h1 = 28.9 m, h2 = 13.1 m, v1 = 0) KE = Kinetic Energy, PE = Potential Energy}

0 + mgh₁ = .5(m)v₂² + mgh₂

(221)(9.8)(28.9) = .5(221)v₂² + (221)(9.8)(13.1)

v₂² = 309.7 v₂ = 17.6 m/s

Change in potential energy is PE2 - PE1 = 34,219.6 N-m