Question

 An aluminum block with mass m was nudged, it slides down the inclined surface at a constant speed when the angle of inclination is 22⁰. Determine the coefficient of kinetic friction between the aluminum block and the inclined surface.

Answer 1

Here $\mathbf{mg}$ is the gravity force, $\mathbf{F_{fr}}$ is the friction force and $\mathbf{N}$ is the reactance,

Since the block is moving with the constant speed, its acceleration is zero. Thus, writting down Newton's second law in the projections on axes, obtain

x-axis:

$$mg\sin\theta - F_{fr} = 0 \\ F_{fr} = mg\sin\theta$$

On the other hand,

$$F_{fr} = \mu N$$

where $\mu$ is the coefficient of kinetic friction between the aluminum block and the inclined surface.

Thus, get:

$$\mu N = mg\sin\theta\\ \mu = \dfrac{mg\sin\theta}{N}$$

y-axis:

$$N-mg\cos\theta = 0\\ N = mg\cos\theta$$

Substituting $N$ into the $\mu$, obtain:

$$\mu = \dfrac{mg\sin\theta}{mg\cos\theta } = \tan\theta\\ \mu = \tan22\degree \approx 0.404$$

Answer. 0.404.