Question

A roller coaster with a mass of 500.0 kg starts from a height of 27.0 meters with a speed of 5.6 m/s. It descends to a height of 2.00 meters before it starts back up. Some energy is loss to the work due to friction. The speed at the bottom of the ramp is 22.0 m/s. How much energy was lost to friction?

Answer 1

The energy loss due to friction is 9465 J.

Step-by-step 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 =
`(1)/(2) * m * v^(2) + m * g* h` =
`(1)/(2) * 500 * 5.6^(2) + 500* 9.81 *27` = 7840 + 132435

E₁ = 140,275 J

The energy of roller at a height 2 m.

E₂ = K.E + P.E

E₂ =
`(1)/(2) * m * v^(2) + m * g* h` =
`(1)/(2) * 500 * 22^(2) + 500* 9.81 *2` = 121000 + 9810

E₂ = 130,810 J

The energy loss due to friction = E₁ - E₂ = 140,275 - 130,810 = 9465 J.