Question

Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above the reference point. If the mass of the car is 1000-kg, and the car traveled a distance of 400 m, estimate the magnitude of the frictional force between the car and the track.

Answer 1

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

$U_{g,1} = U_{g,2} + W_{dis}$

Where:

$U_{g,1}$, $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$W_{dis}$ - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

$U = m \cdot g \cdot y$

Where:

$m$ - Mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$y$ - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

$W_{dis} = f \cdot \Delta s$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

$m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s$

$f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}$

If $m = 1000\,kg$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{1} = 40\,m$, $y_{2} = 25\,m$ and $\Delta s = 400\,m$, then:

$f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}$

$f = 367.763\,N$

The magnitude of the frictional force between the car and the track is 367.763 N.