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You are testing a new amusement park roller coaster with an empty car with a mass of 130 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 . Part A As the car rolls from point A to point B, how much work is done by friction

Sagot :

Answer:

work done by friction = 5889 J

Explanation:

We are given;

Mass of car; m = 130 kg

Speed at point A; v1 = 25 m/s

Speed at point B: v2 = 8 m/s

Since radius is 12 m

At point A, distance is; y1 = 12 m

At point B, distance is; y2 = -12 m

Now, formula for work done by all the forces is given by the equation;

Total work;

W_gravity + W_others = K2 - K1

Where W_others is work done by other forces which is equal to work done by friction

Where K2 - K1 is change in kinetic energy.

W_grav is also change in potential energy and is expressed as;

W_grav = mgy1 - mgy2

K2 - K1 = ½m(v1)² - ½m(v2)²

Thus;

mgy1 - mgy2 + W_others = ½m(v1)² - ½m(v2)²

Making W_others the subject;

W_others = ½m(v1)² - ½m(v2)² + mgy2 - mgy1

Plugging in the relevant values;

W_others = (½ × 130 × 25²) - (½ × 130 × 8²) + (130 × 9.8 × -12) - (130 × 9.8 × 12)

W_others = 5889 J

Recall that I earlier said W_others = work done by friction.

Thus, work done by friction = 5889 J