You are testing a new amusement park roller coaster with an empty car with a mass 120 kg. One part of the track is a vertical loop with radius 12.0 m. At the bottom of the loop(part A) the car has a speed 25.0 m/s and at the top of the loop (point B) has a speed of 8.0 m/s. As the car rolls from point A to point B, how much work is done by friction.

Respuesta :

Answer:

- 5436 J

Explanation:

mass of car, m = 120 kg

radius of loop, r = 12 m

velocity at the bottom (A) = Va = 25 m/s

Velocity at the top(B) = Vb = 8 m/s

Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m

by use of Work energy theorem

Work done by all the forces = change in kinetic energy of the body

Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A

- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)

- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)

- 28224 + Work done by friction = - 33660

Work done by friction = -33660 + 28224 = - 5436 J

ACCESS MORE