Question

A uniform solid ball rolls smoothly along a floor, then up a ramp inclined at 15.2 o. It momentarily stops when it has rolled 2.00 m along the ramp. What was its initial speed?

Answer 1

Answer:

$v \approx 3.207\,\frac{m}{s}$

Explanation:

Let assume that ball's motion is frictionless and its geometric dimensions can be neglected. Besides, it is assumed that ball begins at a height of zero. The ball can be modelled by means of the Principle of Energy Conservation:

$K_{A} = U_{g}$

$\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g \cdot h$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2\,m)\cdot \sin 15.2^{\textdegree}}$

The initial speed of the ball is:

$v \approx 3.207\,\frac{m}{s}$