Question

re testing a new amusement park roller coaster with an empty car with a mass of 100 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 . You may want to review (Pages 203 - 212) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of A vertical circle with friction. Part A As the car rolls from point A to point B, how much work is done by friction?

Answer 1

-4530 J

Step-by-step explanation:

Given that

Mass of the car, m = 100 kg

Speed of the car at point A, v1 = 25 m/s

Speed at point B, v2 = 8 m/s

Radius of the track, r = 12 m and with respect to the origin of the center of the track, we say that y1 = -12 at point A and y2 = 12 at point B

We also know that

W(total) = W(grav) + W(other) = K₂ - K₁

Work done by the gravitational force, W(grav) = -U(grav) = mgy1 - mgy2

Kinetic Energy, K = ½mv²

Adding all together, we have

½mv₁² + mgy1 + W(other) = ½mv₂² + mgy2

½ * 100 * 25² + 100 * 9.8 * -12 + W = ½ * 100 * 8² + 100 * 9.8 * 12

50 * 625 + 980 * -12 + W = 50 * 64 + 980 * 12

31250 - 11760 + W = 3200 + 11760

19490 + W = 14960

W = 14960 - 19490

W = -4530 J