Answer to Question #129681 in Physics for SCORI

Question

a solid ball of radius R and mass m is pushed with an initial velocity of 3 m / s onto an inclined plane of 30 degrees with a kinetic friction coefficient of 0.2 and a coefficient of static friction of 0.3. When climbing the inclined plane, the ball always rolls and never slips. The maximum height the ball can reach is approx

Answer 1

According to the law of conservation of energy, the change in kinetic and potential energy is equal to the work done by friction force.

$$\Delta T + \Delta U = W_f$$

Let $h$ be the maximum height the ball can reach. At this height ball is in rest and the change in kinetic energy is:

$$\Delta T = 0-\dfrac{mv_0^2}{2 } = -\dfrac{mv_0^2}{2 }$$

where $v_0$ is a initial velocity.

The change in potential energy is:

$$\Delta U = mgh - 0 = mgh$$

The work done by friction force is:

$$W_f = -F_f\cdot \dfrac{h}{\cos \theta}$$

where $\theta = 30\degree.$ and "-" sign is because the force is directed in an opposite direction with respect to displacement.

The friction force for the rolling ball is:

$$F_f = \dfrac{kmg}{R}$$

where $k = 0.2 m$ is a kinetic friction coefficient.

Combining it all together, obtain:

$$\Delta T + \Delta U = W_f\\ -\dfrac{mv_0^2}{2 } + mgh = -\dfrac{kmg}{R}\cdot \dfrac{h}{\cos \theta}$$

Expressing $h$, get:

$$h = \dfrac{v^2_0}{2g\left ( 1 + \frac{k}{R\cos\theta}\right )}$$

Answer. $h = \dfrac{v^2_0}{2g\left ( 1 + \frac{k}{R\cos\theta}\right )}$.