Question

A particle of mass M is released from rest on a rough inclined plane, which makes an
angle of 30° with the horizontal. It is observed that the particle moves a distance of 3 m
in 3 s. What is the particle’s acceleration? Draw a properly labelled free body diagram.
Calculate the coefficient of kinetic friction between the particle and the surface of the
inclined plane.

Answer 1

Answer on Question 61693, Physics, Mechanics, Relativity

Question:

A particle of mass $M$ is released from rest on a rough inclined plane, which makes an angle of $30{}^{\circ}$ with the horizontal. It is observed that the particle moves a distance of $3\,m$ in $3\,s$. What is the particle’s acceleration? Draw a properly labelled free body diagram. Calculate the coefficient of kinetic friction between the particle and the surface of the inclined plane.

Solution:

a) We can find the particle’s acceleration from the kinematic equation:

$$d = v_0 t + \frac{1}{2} a t^2,$$

here, $d$ is the distance, $v_0$ is the initial velocity of the particle (because the particle is released from rest $v_0 = 0\,ms^{-1}$), $t$ is the time during which the particle moved the distance $d$ and $a$ is the particle’s acceleration which we are searching for.

Then, from this formula we can find the particle’s acceleration:

$$d = \frac{1}{2} a t^2,$$ $$a = \frac{2d}{t^2} = \frac{2 \cdot 3\,m}{(3\,s)^2} = 0.67\, \frac{m}{s^2}.$$

b) There are three forces that act on the particle: the force of gravity $Mg$ directed downward and can be resolved into two perpendicular components ($F_{\parallel} = M g \sin \theta$ and $F_{\perp} = M g \cos \theta$), the force of reaction directed perpendicular to the surface of the inclined plane and the friction force $F_{fr}$ directed opposite to the motion of the particle. Let’s draw a free-body diagram and write all the forces that act on the particle:

$$M \vec {g} + \vec {N} + \overrightarrow {F _ {f r}} = m \vec {a}.$$

Then projected the forces on axis $x$ and $y$ we get:

$$M g \sin \theta - F _ {f r} = M a,$$ $$N - M g \cos \theta = 0.$$

Let's find the friction force that acts on the particle:

$$F _ {f r} = \mu_ {k} N = \mu_ {k} M g \cos \theta .$$

Substituting the friction force into the first equation we get:

$$M g \sin \theta - \mu_ {k} M g \cos \theta = M a,$$ $$g \sin \theta - \mu_ {k} g \cos \theta = a.$$

From the last equation we can find the coefficient of kinetic friction between the particle and the surface of the inclined plane:

$$\mu_ {k} = \frac {g \sin \theta - a}{g \cos \theta} = \frac {9 . 8 \frac {m}{s ^ {2}} \cdot \sin 3 0 {}^ {\circ} - 0 . 6 7 \frac {m}{s ^ {2}}}{9 . 8 \frac {m}{s ^ {2}} \cdot \cos 3 0 {}^ {\circ}} = 0. 5.$$

Answer:

a) $a = 0.67\frac{m}{s^2}.$

b) $\mu_{k} = 0.5$

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