91Ó°ÊÓ

Potential energy: Potential energy is stored in body due to the virtue of its position.If a body of mass m is raised against gravity g to a certain height h, the potential energy stored in the body is,

$\mathrm{Δ}\text{PE}=mgh$

Kinetic energy: Kinetic energy is stored in body due to the virtue of its motion.If a body of mass m is moving with a velocity of v, the kinetic energy stored in the body is,

$\mathrm{Δ}\text{KE}=\frac{1}{2}m{v}^2$

The potential energy associated with the roller coaster is,

$\mathrm{Δ}\text{PE}=mgh$

Here, m is the mass of the roller coaster, g is the acceleration due to gravity $\left(9.8\text{m}/{\text{s}}^2\right)$, and h is the height descended by the roller coaster$\left(h=20.0\text{m}\right)$.

The initial kinetic energy associated with the roller coaster is,

$\mathrm{Δ}{\text{KE}}_{\text{i}}=\frac{1}{2}m{v}_i^2$

Here, m is the mass of the roller coaster, and$v_i$is the initial velocity of the roller coaster$\left(v_i=5.0\text{m}/\text{s}\right)$.

The ratio of the potential energy to the initial kinetic energy is,

$\begin{array}{rcl}\frac{\mathrm{Δ}\text{PE}}{\mathrm{Δ}{\text{KE}}_{\text{i}}}& =& \frac{mgh}{\frac{1}{2}m{v}_i^2}\\ & =& \frac{gh}{0.5v_i^2}\end{array}$

Putting all known values,

$\begin{array}{rcl}\frac{\mathrm{Δ}\text{PE}}{\mathrm{Δ}{\text{KE}}_{\text{i}}}& =& \frac{\left(9.8\text{m}/{\text{s}}^2\right)×\left(20.0\text{m}\right)}{0.5×{\left(5.0\text{m}/\text{s}\right)}^2}\\ & =& 15.68\end{array}$

As a result, the ratio of the potential energy to kinetic energy is 15.68.

Therefore, the potential energy is much greater than initial kinetic energy.