A snowboarder coasts on a smooth track that rises from one level to another, If the snowboarder's...

Question:

A snowboarder coasts on a smooth track that rises from one level to another, If the snowboarder's initial speed is 4 m/s, the snowboarder just makes it to the upper level and comes to rest. With a slightly greater initial speed of 5 m/s, the snowboarder is still moving to the right on the upper level. His final speed, in this case, is 3 m/s. Suppose this situation is repeated on the fictional planet Epsilon, where the acceleration due to gravity is less than it is on the Earth.

a. Would the height of a hill on Epsilon that causes a reduction in speed from 4 m/s to 0 be greater than, less than, or equal to the height of the corresponding hill on Earth? Explain.

b. Consider the hill on Epsilon discussed in part a. lf the initial speed at the bottom of the hill is 5 m/s, will the final speed at the top of the hill be greater than, less than, or equal to 3 m/s? Explain.

Conservation of Energy:

When there are no outside forces acting on a system, the total amount of energy in that system must remain constant. The energy can change form (from gravitational potential to kinetic, for example), but the total amount of energy must remain constant. The equations fr gravitational potential energy, Eg and kinetic energy, Ek follow:

{eq}E_g=m*g*h\\ E_k=\frac{1}{2}*m*v^2 {/eq}

m is the mass

g is the acceleration due to gravity

h is the height

m is the mass

v is the speed

If we do the same experiment on two different planets, the gravitational acceleration g is likely different and, in which the mechanical energy is conserved in both cases, that changes the maximum height and/or the speed of the snowboarder.

Part (a)

The height reached by the snowboarder on Epsilon is greater than the height the snowboarder reaches on earth.

For the gravitational energy {eq}g {/eq} on earth, the snowboarder (starting at speed v) makes it to the top level at rest, meaning that the mechanical energy at that point is mgh i.e. entirely gravitational potential energy. The equation would be:

{eq}\begin{align*} \text{kinetic energy Lost} &= \text{Gravitational potential energy Gained}\\ \frac{1}{2}mv^2 &= mgh\\ \end{align*} {/eq}

For this case, the mass m is the same.

For the height {eq}h_E {/eq} and gravitational acceleration {eq}g_E \ (< g) {/eq} on Episilon, the mechanical energy should be the same, i.e.

{eq}mgh = mg_Eh_E {/eq}

For us to have a balance, the height {eq}h_E {/eq} should overcompensate for the RHS, since the gravitational acceleration {eq}g_E \ (< g) {/eq} i.e. {eq}\color{blue}{\boxed{h_E > h}} {/eq}

Part b

For the case on earth, where the initial speed is u = 5 m/s, the final speed is v = 3 m/s and the height is h,

{eq}\begin{align*} \text{Initial Kinetic energy} &= \text{Total mechanical Energy}\\ \frac{1}{2}mu^2 &= \frac{1}{2}mv^2+ mgh \end{align*} {/eq}

In our investigation for this case, the height h and the mass m are the same.

On episilon, where the graviational acceleration is less,

{eq}\begin{align*} \frac{1}{2}mu^2 = \frac{1}{2}mv^2+ mgh &= \frac{1}{2}mv_E^2+ mg_Eh \end{align*} {/eq}

Since {eq}g_E \ (< g) {/eq}, the part {eq}mg_Eh < mgh {/eq} and thus the part {eq}\frac{1}{2}mv_E^2 {/eq} has to overcompensate, such that:

{eq}\begin{align*} \frac{1}{2}mv_E^2 &> \frac{1}{2}mv^2\\ v_E^2 &> v^2\\ v_E &> v\\ \color{blue}{v_E} &> \color{blue}{3\ \text{m/s}} \end{align*} {/eq}