Answer :
Answer:
[tex]33\,000\; {\rm J}[/tex].
Approximately [tex]20\; {\rm m}[/tex] (assuming that [tex]g = 10\; {\rm m\cdot s^{-2}}[/tex], and that kinetic energy is entirely converted into gravitational potential energy.)
Explanation:
When an object of mass [tex]m[/tex] travels at a speed of [tex]v[/tex], the kinetic energy [tex]({\rm KE})[/tex] of that object would be:
[tex]\displaystyle ({\rm KE}) = \frac{1}{2}\, m\, v^{2}[/tex].
In this question:
- [tex]m = 165\; {\rm kg}[/tex].
- [tex]v = 20\; {\rm m\cdot s^{-1}}[/tex].
Therefore, kinetic energy would be:
[tex]\begin{aligned} ({\rm KE}) &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\, (165\; {\rm kg})\, (20\; {\rm m\cdot s^{-1}})^{2} \\ &= 33\, 000\; {\rm J} \end{aligned}[/tex].
Under the assumption that kinetic energy [tex](\text{KE})[/tex] of the roller coaster is entirely converted into gravitational potential energy [tex](\text{GPE})[/tex], the maximum possible increase in the [tex](\text{GPE})[/tex] of roller coaster should be equal to the initial [tex](\text{KE})[/tex], [tex]33\, 000\; {\rm J}[/tex].
When an object of mass [tex]m[/tex] gains a height of [tex]h[/tex], the gain in [tex](\text{GPE})[/tex] would be:
[tex](\text{GPE}) = m\, g\, h[/tex],
Where [tex]g = 10\; {\rm m\cdot s^{-2}}[/tex] is the gravitational field strength under the assumptions. Given that the gain in [tex](\text{GPE})[/tex] is [tex](1/2)\, m\, v^{2} = 33\, 000\; {\rm J}[/tex] (the maximum possible value), rearrange the equation above to find the increase in the height of the roller coaster:
[tex]\displaystyle m\, g\, h = \frac{1}{2}\, m\, v^{2}[/tex].
[tex]\begin{aligned} h &= \frac{(1/2)\, m\, v^{2}}{m\, g} \\ &= \frac{v^{2}}{2\, g} \\ &= \frac{(20\; {\rm m\cdot s^{-1}})^{2}}{(2)\, (10\; {\rm m\cdot s^{-2}})} \\ &= 20\; {\rm m}\end{aligned}[/tex].
In other words, the maximum possible height gain of the roller coaster would be [tex]20\; {\rm m}[/tex] under these assumptions.
