Answer to Question #234576 in Physics for mjh

Question

A roller coaster reaches the top of the steepest hill with a speed of 6.8 km/h

km/h . It then descends the hill, which is at an average angle of 39∘ and is 37.0 m long. What will its speed be when it reaches the bottom? Assume μk = 0.18.

Answer 1

According to conservation of energy, the initial mechanical energy (potential energy plus kinetic energy) will be converted into kinetic energy:

$$E_1=mgh+\frac12mv^2=\\\space\\ =mgL\sin\theta+\frac12mv^2.$$

This mechanical energy will be spent to overcome friction and gain speed while moving downward:

$$E_2=\mu mgL\sin\theta+\frac12mu^2.$$

By conservation of energy:

$$E_1=E_2,\\\space\\ mgL\sin\theta+\frac12mv^2=\mu mgL\sin\theta+\frac12mu^2,\\\space\\ gL\sin\theta+\frac12v^2=\mu gL\sin\theta+\frac12u^2,\\\space\\ u=\sqrt{2gL\sin\theta(1-\mu)+v^2}=20.5\text{ m/s}.$$