Question

a small block starts sliding down an inclined plane forming an angle with the horizontal the coefficient of friction depends on the distance covered from rest along as ax where a=constant. find distance covered by the block down the plane, till it stops sliding,and its max velocity during this journey

Answer 1

Answer on Question#39296, Physics, Mechanics | Kinematics | Dynamics

A small block starts sliding down an inclined plane forming an angle with the horizontal the coefficient of friction depends on the distance covered from rest along as ax where $a =$ constant. find distance covered by the block down the plane, till it stops sliding, and its max velocity during this journey.

Solution:

To calculate the forces on an object placed on an inclined plane, consider the three forces acting on it.

- The normal force (N) exerted on the body by the plane due to the attraction of gravity i.e. $\mathrm{N} = \mathrm{{mg}}\cos \theta$

the force due to gravity (mg, acting vertically downwards) $F =$ mgsin $\theta$

- the frictional force (f) acting parallel to the plane. $f = \mu N = \mu$ mg cos $\theta$ , where $\mu$ is the coefficient of friction $\mu = a_{k} x$ ( $a_{k} =$ constant).

Equation of motion:

$$m a = m g \sin \theta - f = m g \sin \theta - \mu m g \cos \theta = m g (\sin \theta - a _ {k} x \cos \theta)$$

Acceleration

$$a = g (\sin \theta - a _ {k} x \cos \theta)$$ $$a = \frac {d v}{d t} = \frac {d v}{d x} \frac {d x}{d t} = v \frac {d v}{d x}$$ $$v \frac {d v}{d x} = g (\sin \theta - a _ {k} x \cos \theta)$$ $$v d v = g (\sin \theta - a _ {k} x \cos \theta) d x$$ $$\int_ {0} ^ {v} v d v = \int_ {0} ^ {x} g (\sin \theta - a _ {k} x \cos \theta) d x$$ $$\frac {v ^ {2}}{2} = g x \sin \theta - a _ {k} g \frac {x ^ {2}}{2} \cos \theta$$

Block stops when velocity $v = 0$ and distance $x = d$ covered by the block down the plane

$$g d \sin \theta - a _ {k} g \frac {d ^ {2}}{2} \cos \theta = 0$$ $$d = \frac {2 g \sin \theta}{a _ {k} g \cos \theta} = \frac {2}{a _ {k}} \tan \theta$$

The max velocity we will find by taking the derivative of function and equal it to zero

$$\frac {v ^ {2}}{2} = f (x) = g x \sin \theta - a _ {k} g \frac {x ^ {2}}{2} \cos \theta$$ $$f ^ {\prime} (x) = \frac {d}{d x} \left(g x \sin \theta - a _ {k} g \frac {x ^ {2}}{2} \cos \theta\right) = g (\sin \theta - a _ {k} x \cos \theta) = 0$$

The max velocity will be at point

$$x = \frac {\sin \theta}{a _ {k} \cos \theta} = \frac {\tan \theta}{a _ {k}}$$ $$\frac {v _ {m a x} ^ {2}}{2} = f \left(\frac {\tan \theta}{a _ {k}}\right) = g \frac {\tan \theta}{a _ {k}} \sin \theta - a _ {k} g \frac {\left(\frac {\tan \theta}{a _ {k}}\right) ^ {2}}{2} \cos \theta = \frac {g \sin \theta \tan \theta}{2 a _ {k}}$$

Thus,

$$v _ {m a x} = \sqrt {\frac {g \sin \theta \tan \theta}{a _ {k}}}$$

Answer. 1. $d = \frac{2}{a} \tan \theta$ ; 2. $v_{max} = \sqrt{\frac{g \sin \theta \tan \theta}{a}}$ .