Question

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?

Answer 1

`W_f=-62460\ J`

Step-by-step 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

`m g * (2R) + W_f=(1)/(2)mv^2-(1)/(2)mu^2`

Now by putting the values in the above equation we get

`120* 10* 2* 12+ W_f=(1)/(2)* 120* 8^2-(1)/(2)* 120* 25^2`

`W_f=(1)/(2)* 120* 8^2-(1)/(2)* 120* 25^2-120* 10* 2* 12\ J`

`W_f=-62460\ J`

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