# A roller coaster cart with mass m= 150 kg is at the top of a ride travelling at v_0 = 10 m/s on frictionless tracks. The cart then descends on the almost vertical drop of the ride. The height is 110 m high. What is the cart's speed at base of the ride?

Nov 8, 2015

$47.5 \text{m/s}$

#### Explanation:

I will apply the conservation of energy. The sum of the kinetic and potential energy at the top of the ride will be converted into kinetic energy at the bottom:

$\frac{1}{2} m {v}_{0}^{2} + m g h = \frac{1}{2} m {v}^{2}$

$\therefore \frac{1}{2} \times 150 \times {10}^{2} + \left(150 \times 9.8 \times 110\right) = \frac{1}{2} \times 150 \times {v}^{2}$

$\therefore 7500 + 161700 = 75 {v}^{2}$

$\therefore v = \sqrt{2256}$

$v = 47.5 \text{m/s}$