Question

You are testing a new amusement park roller coaster with an empty car with a mass of 108kg. One part of the track is a vertical loop with a radius of 12.0m. At the bottom of the loop (pointA) the car has a speed of 25.0m/sand at the top of the loop (pointB) it has speed of 8.00m/s.As the car rolls from point A to point B, how much work is done by friction?

Answer 1

Answer:

-4892 J

Explanation:

The work done by friction is equal to the difference in mechanical energy of the car between the bottom of the loop (point A) and the top of the loop (point B).

The mechanical energy of the car at point A is

$E_A = \frac{1}{2}mv_A^2 + mgh_A$

where

$m=108 kg$ is the mass of the car

$v_A = 25.0 m/s$ is the speed at point A

$h_A=0$ is the height of the car at point A (zero because it is at the bottom of the loop)

Substituting into the equation, we find

$E_A = \frac{1}{2}(108 kg)(25.0 m/s)^2 + (108 kg)(9.8 m/s^2)(0)=33,750J$

The mechanical energy of the car at point B is

$E_B = \frac{1}{2}mv_B^2 + mgh_B$

where

$m=108 kg$ is the mass of the car

$v_B = 8.0 m/s$ is the speed at point B

$h_B=24.0 m$ (twice the radius) is the height of the car at point B, at the top of the loop

Substituting into the equation, we find

$E_B = \frac{1}{2}(108 kg)(8.0 m/s)^2 + (108 kg)(9.8 m/s^2)(24.0 m)=28,858J$

So, the work done by friction is

$W=E_B-E_A=28,858 J-33,750 J=-4,892 J$

And the work is negative because it is done against the motion of the car.