Question

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically drops that distance before the track curves forward. If the velocity at the bottom of the drop is 10.0 m/s and the mass of the cart and passengers is 3.5 x 104 kg, how much potential energy was converted into thermal energy (heat)?

3.7 x 107 J
4.2 x 107 J
4.5 x 107 J
8.0 x 107 J

Answer 1

Answer:

Potential energy converted = 4.2*10^7 [J]

Explanation:

This problem has to be analysed under the principle of energy conservation. For this case potential energy that is transformed into kinetic energy, but however not all potential energy is transformed into kinetic energy, as we are told that some of this energy dissipates in the form of heat. We must first find the potential energy.

Potential energy

$E_{p} =m*g*h\\where:\\m = mass =3.5*10^4[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 126.5 [m]\\therefore\\E_{k} = 3.5*10^4*9.81*126.5\\E_{k} = 43433775 [J]$

Kinetic Energy

$E_{k}=\frac{1}{2} * m*v^{2}\\ where:\\v = velocity = 10[m/s]\\therefore:\\E_{k}=\frac{1}{2} * 3.5*10^4*10^{2}\\\\E_{k}=1750000[J]$

As we can see, not all potential energy has been transformed into kinetic energy. That is, part of this energy dissipates as thermal energy, so this difference of energies will be equal to the loss of energy represented as dissipated energy.

$Thermal Energy\\T_{e} = 43433775-1750000\\T_{e} =41683775[J]$

The previous value can be approximated to 4.2 * 10^7 [J]