As the car rolls from point A to point B, how much work is done by friction?

Question

You are testing a new amusement park roller coaster with an empty car with a mass of 108 kg. One part of the track is a vertical loop with a radius of 12.0 m. At the bottom of the loop (point A) the car has a speed of 25.0 m/s and at the top of the loop (point B) it has speed of 8.00 m/s.

As the car rolls from point A to point B, how much work is done by friction?
Use g = 9.81 m/s^2 for the acceleration due to gravity.

Answer 1

Calculate the kinetic energy (mv²/2) of the car at the top and bottom of the loop. Subtract the Ke at the top from the Ke at the bottom to get the *total* work done. The amount of work done by gravity is mgh so subtract that from the total work done, and the result is the work done by friction.

HTHY. Have fun ☺

Doug

Answer 2

Work done by non-conservative forces = Change in Energy

Since friction is non-conservative (NC), we have

W(friction) = E(final) - E(initial)
W = (1/2 m vf^2 + m g hf) - (1/2 m vi ^2 + m g hi)
W = (.5 * 108 * 8^2 + 108 * 9.8 * 24) - (.5 * 108 * 25^2 + 108*9.8*0)

W = 28857.6 - 33750 J
W = -4892.4 J

The answer is approximately -4890 J or -4.89 kJ.

The answer is negative because friction REMOVES energy from the system.