Question

A roller-coaster car is moving at 20 m/s along a straight horizontal track. What will its speed be after climbing the 15-m hill shown in the figure, if friction is ignored?

http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c06/q40-1.png

Answer 1

A roller-coaster car is moving at $20\,\mathrm{m/s}$ along a straight horizontal track. What will its speed be after climbing the 15-m hill shown in the figure, if friction is ignored?

Solution:

Conservation of energy:

$$W_{\text{kinetic start}} + W_{\text{potential start}} = W_{\text{kinetic top}} + W_{\text{potential at top}}$$

let's call the level of the horizontal track our reference level, so potential energy will be zero there, so we have:

$$\begin{array}{l} \frac{m V_{\text{start}}^2}{2} = \frac{m V_{\text{top}}^2}{2} + mgh \\ V_{\text{top}}^2 = V_{\text{start}}^2 - 2gh \\ \end{array}$$ $$V_{\text{top}} = \sqrt{V_{\text{start}}^2 - 2gh} = \sqrt{\left(20\,\frac{m}{s}\right)^2 - 2 \cdot 9.8\,\frac{m}{s^2} \cdot 15\,\mathrm{m}} = 10.3\,\frac{\mathrm{m}}{\mathrm{s}}$$

Answer: speed of the car will be $10.3\,\frac{\mathrm{m}}{\mathrm{s}}$.