Question

You are testing a new roller coaster ride in which a car of mass m moves around a vertical circle of radius R. In one test, the car starts at the bottom of the circle (point A) with initial kinetic energy Ki. When the car reaches the top of the circle (point B), its kinetic energy is Ki/4, and its gravitational potential energy has increased by Ki/2. How much work was done on the car by the force of friction as it moved from point A to point B, in terms of m, g and R? What was the magnitude of the friction force (assumed to be constant throughout the motion), in terms of m and g?

Answer 1

Answer:

• The work done by friction is equal to mgR.

• The magnitude of the frictional force is mg/π.

Explanation:

From the conservation of mechanical energy, we know that the sum of the kinetic energy, the gravitational potential energy and the work done by the friction must be conserved in all the trajectory. Assuming that the potential energy is zero at the bottom of the circle, we can say that:

$K_i=K_f+U_f+W_f$  (*)

Where $K_f$ is the final kinetic energy, $U_f$ is the final potential energy and $W_f$ is the work done by the friction.

Now, it is said that the final kinetic energy is $\frac{K_i}{4}$, and the final potential energy is $\frac{K_i}{2}$. We know the expression for gravitational potential energy:

$U=mgh$

Since the height h of the car at the top of the circle is 2R, we can write the final potential energy as:

$U_f=mg(2R)=2mgR$

And then, the initial kinetic energy is:

$U_f=\frac{K_i}{2}=2mgR\\\\\implies K_i=4mgR$

Now we can rewrite the equation (*) as:

$4mgR=\frac{4mgR}{4}+2mgR+W_f\\ \\\implies W_f=mgR$

So, the work done on the car by the friction is mgR.

Then, since work is defined as force times distance (when the force is constant) we just need to obtain the distance in which the friction acted on the car. This is the length of the semicircle of radius R:

$s=\frac{2\pi R}{2}=\pi R$

Finally, we solve the frictional force from the definition of work:

$W_f=fs\\\\f=\frac{W_f}{s}\\\\f=\frac{mgR}{\pi R}\\ \\f=\frac{mg}{\pi}$

It means that the magnitude of the frictional force is mg/π.