Physics, energy problems

Question

1) a 2000 kg rollercoaster starts from a height of 55 m and an initrial velocity of 8 m/s. When it reaches the bottom of the hill (height of zero), what is its velocity?

2) For the rollercoaster problem above (m=2000kg), if the final velocity at the bottom of the hill was only 27 m/s instead of 34 m/s, how much energy would have been "lost"? (Lost mechanical energy is the Eother)

Answer 1

Using the concept of conservation of energy and assuming no lost of other energy, the change in potential energy, ΔP is equal to the (negative) change in kinetic energy, - ΔKE

==> mg(y₂ - y₁) = -.5m(v₂² - v₁²), where y₂, y₁ are the initial and final heights respectively and v₁ and v₂ are the initial and final speed respectively.

a) Solving for the final speed yields

v₂ = √(gy₂ +.5v₁2) = 34m/s

NB: total energy, E = ΔKE +ΔPE = 1092000J + 1078000J = 2170000J

b) the total energy E = ΔKE +ΔPE +E_other

If v₂ = 27m/s, the change in kinetic energy ΔKE = 665000J

==> E_other = E - ΔKE - ΔPE = 2170000J - 665000J - 1078000J = 427000J

Notice that this result is same as finding the difference in the change in kinetic between part(a) and part(b)

i.e E_other = 1092000J - 665000J = 427000J