Question

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

Answer 1

Answer:

W = - 23.309 KJ

Explanation:

m= 120 kg

Vi= 25 m/s

Vf= 8 m/s

h= 12 m

From energy conservation

Lets take work done by friction is W

W(friction) = E(final) - E(initial)

W = (1/2 m Vf² + m g hf) - (1/2 m Vi² ^2 + m g hi)

Now by putting the values

W = (0.5 x 120 x 8² + 120 x 9.81 x 12) - (0.5 x 120 x 25²  + 120 x 9.81 x 0)

W = -23309.6 J

W = - 23.309 KJ

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