Question

In Example 7.7, we found that the speed of a roller coaster that had descended 20.0m was only slightly greater when it had an initial speed of 5.00m/s than when it started from rest. This implies that $\mathit{\Delta }\mathbf{\text{PE}}\mathbf{\gg }\mathit{\Delta }{\mathbf{\text{KE}}}_{\mathbf{\text{i}}}$.Confirm this statement by taking the ratio of $\mathit{\Delta }\mathbf{\text{PE}}$to $\mathit{\Delta }{\mathbf{\text{KE}}}_{\mathbf{\text{i}}}$. (Note that mass cancels.)

Short answer

The ratio of the potential energy to kinetic energy is 15.68. Therefore, the potential energy is much greater than initial kinetic energy.

Step-by-step solution

Potential and Kinetic energy

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,

$\Delta \text{PE}=mgh$

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,

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

Ratio of potential energy to the initial kinetic energy

The potential energy associated with the roller coaster is,

$\Delta \text{PE}=mgh$

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,

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

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,

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

Putting all known values,

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

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.