# A 500 kg roller coaster starts from rest at points 35 m above the ground level. The car drives...

## Question:

A 500 kg roller coaster starts from rest at points 35 m above the ground level. The car drives down into a valley 4 m above the ground and then climbs the top of a hill that is 24 m above the ground level. What is the velocity of the car in the valley and at the top of the hill? (Neglect friction)

## Energy Conservation Principle:

The energy conservation principle consists of a physical law that states that if no external forces act on the system, the total energy of the system should conserve. Energy is not created nor destroyed, it only transforms. This means that a system may present energy transformations, but the overall total energy of the system will remain the same. The total energy of a system can be determined by the sum of the total kinetic and potential energy of the system.

$$E_t=K_e+U_e\\ E_t=0.5mv^2+mgh$$

In order to determine the velocity at the different points in the roller coaster, we must use energy conservation mechanics. We take the initial position, and then determine the amount of kinetic and potential energy in the valley.

{eq}E_i=E_f\\ U_e=K_{ef}+U_{ef}\\ mgh_1=\dfrac{1}{2}mv_1^2+mgh_2\\ 2gh_1-2gh_2=mv_1^2\\ v_1=\sqrt{\dfrac{2gh_1-2gh_2}{m}}\\ v_1=\sqrt{\dfrac{2\times9.8\times35-2\times9.8\times4}{500}}\\ v_1=1.10\ m/s {/eq}

At the top of the hill:

{eq}E_i=E_f\\ U_e=K_{ef}+U_{ef}\\ mgh_1=\dfrac{1}{2}mv_1^2+mgh_2\\ 2gh_1-2gh_2=mv_2^2\\ v_2=\sqrt{\dfrac{2gh_1-2gh_2}{m}}\\ v_2=\sqrt{\dfrac{2\times9.8\times35-2\times9.8\times24}{500}}\\ v_2=0.65\ m/s {/eq}