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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.

User Jimadine
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2 Answers

5 votes

Final answer:

The work done by friction as the empty roller coaster car moves from the bottom to the top of the loop is found using the change in kinetic energy. With a speed of 25.0 m/s at the bottom and 8.0 m/s at the top, the work done by friction is calculated to be -27,360 joules.

Step-by-step explanation:

moves from the bottom to the top of a vertical loop in an amusement park. We have the car's mass (120 kg), its speed at the bottom of the loop (25.0 m/s), and its speed at the top of the loop (8.0 m/s).

The change in kinetic energy (ΔKE) from point A to B is given by:

ΔKE = KEtop - KEbottom = ½ m vB2 - ½ m vA2

Where:

  • m is the mass of the car (120 kg)
  • vA is the speed at the bottom of the loop (25.0 m/s)
  • vB is the speed at the top of the loop (8.0 m/s)

Plugging in the values:

ΔKE = ½ × 120 kg × (8.0 m/s)2 - ½ × 120 kg × (25.0 m/s)2

= -27,360 J (since the final kinetic energy is less than the initial, the change is negative, indicating a loss of energy due to work done against friction and other forces)

This means the work done by friction, along with any other non-conservative forces (if present), as the car moves from point A to B is -27,360 joules.

User Denziloe
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4.8k points
5 votes

Answer:

- 5436 J

Step-by-step 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

User Rohit Jangid
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4.7k points