Question

Given an inclined plane with a slope of 1 in 4 measured along the plane. A metal block of mass 2 kg. slides uniformly. What is the coefficient of friction?

Answer 1

Answer on Question #64164, Physics / Mechanics | Relativity

Question:

Given an inclined plane with a slope of 1 in 4 measured along the plane. A metal block of mass $2\mathrm{kg}$ slides uniformly. What is the coefficient of friction?

Solution:

There are two forces acting on this block: its weight and the force of friction. Because the block slides uniformly we may conclude that its acceleration is zero and the magnitude of vector $\vec{F}_{fr}$ is equal to the magnitude of vector $\vec{F}_T$ (tangential component of weight).

Let $m$ be the mass of the block, $k$ — coefficient of friction, $g$ — acceleration of gravity.

$$\left| \vec {F} _ {f r} \right| = k \left| \vec {F} _ {P} \right| = k m g \cos \alpha$$ $$\left| \vec {F} _ {T} \right| = m g \sin \alpha$$ $$\left| \vec {F} _ {f r} \right| = \left| \vec {F} _ {T} \right| \Rightarrow k m g \cos \alpha = m g \sin \alpha \Rightarrow k = \tan \alpha$$

The value of tan $\alpha$ is equal to the slope of our plane, that is $\frac{1}{4}$ . So, $= \frac{1}{4} = 0.25$ .

Answer:

0.25

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