You are testing a new amusement park roller coaster with an empty car with a mass of 120 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?

Explanation:

Given that

mass of the car ,m = 120 kg

Radius ,R= 12 m

Speed at the bottom , u = 25 m/s

Speed at top ,v= 8 m/s

We know that

Work done by all the forces = Change in the kinetic energy

Work done by gravity +  Work done by friction =Change in the kinetic energy

By taking point A as reference

Now by putting the values in the above equation we get

Therefore the work done by friction force will be -62460 J.

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

3.2x10^7