Question

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves down a frictionless track to a height of 3.00 m. How fast are you moving when you arrive at the 3.00-m height?

Answer 1

Answer: 20.765 m/s

Explanation:

This problem can be solved by the conservation of energy principle, this means the initial energy $E_{o}$ must be equal to the final energy  $E_{f}$:

$E_{o}=E_{f}$ (1)

Where each energy is the sum of kinetic energy $K$ and potential energy $U$:

$K_{o}+U_{o}=K_{f}+U_{f}$ (2)

Where:

$K_{o}=\frac{1}{2}mV_{o}^{2}$

Being $m$ your mass and $V_{o}=0 m/s$ your initial velocity, since the roller coaster sterted from rest.

$U_{o}=mgh_{o}$

Being  $g=9.8 m/s^{2}$ the acceleration due gravity and  $h_{o}=25 m$ your initial height

$K_{f}=\frac{1}{2}mV_{f}^{2}$

Being $V_{f}$ your final velocity

$U_{f}=mgh_{f}$

Being $h_{f}=3 m$ your final height

Rewritting (2):

$\frac{1}{2}mV_{o}^{2}+mgh_{o}=\frac{1}{2}mV_{f}^{2}+mgh_{f}$ (3)

$mgh_{o}=m(\frac{1}{2}V_{f}^{2}+gh_{f})$ (4)

Isolating $V_{f}$:

$V_{f}=\sqrt{2g(h_{o}-h_{f})}$ (5)

$V_{f}=\sqrt{2(9.8 m/s^{2})(25 m-3 m)}$ (6)

Finally:

$V_{f}=20.765 m/s$ This is your spedd when you arrive at 3 m height