Answer:
The energy loss due to friction is 9465 J.
Explanation:
The given data :-
Mass of roller (m) = 500 kg.
The speed of roller coaster at height 27 m = 5.6 m/s.
Initial height of roller( h ) = 27 m.
The speed of roller coaster at bottom = 22 m/s.
Initial energy of roller at a height 27 m.
E₁ = K.E + P.E
E₁ = [tex]\frac{1}{2} * m * v^{2} + m * g* h[/tex] = [tex]\frac{1}{2} * 500 * 5.6^{2} + 500* 9.81 *27[/tex] = 7840 + 132435
E₁ = 140,275 J
The energy of roller at a height 2 m.
E₂ = K.E + P.E
E₂ = [tex]\frac{1}{2} * m * v^{2} + m * g* h[/tex] = [tex]\frac{1}{2} * 500 * 22^{2} + 500* 9.81 *2[/tex] = 121000 + 9810
E₂ = 130,810 J
The energy loss due to friction = E₁ - E₂ = 140,275 - 130,810 = 9465 J.
