Friction MCQ Questions & Answers in Basic Physics | Physics

When the body has maximum speed then
$$\eqalign{ & \mu = 0.3x = \tan {45^ \circ } \cr & \therefore x = 3.33\,m \cr} $$

Frictional force on the box $$f = \mu mg$$
$$\therefore $$ Acceleration in the box $$a = \mu g = 5\,m{s^{ - 2}}$$
$${v^2} = {u^2} + 2as \Rightarrow 0 = {2^2} + 2 \times \left( 5 \right)s \Rightarrow s = - \frac{2}{5}$$
w.r.t. belt ⇒ distance $$= 0.4\,cm$$

For the block to remain stationary with the wall

$$\eqalign{ & f = W \cr & \mu N = W \cr & 0.2 \times 10 = W \Rightarrow W = 2N \cr} $$

Friction is the retarding force for the block
$$F = ma = \mu R = \mu mg$$
Therefore, from the first equation of motion
$$\eqalign{ & v = u - at \cr & 0 = V - \mu g \times t \Rightarrow \frac{V}{{\mu g}} = t \cr} $$

When, a plane is inclined to the horizontal at an angle $$\theta ,$$ which is greater than the angle of repose, the body placed on the inclined plane slides down with an acceleration $$a.$$

As, it is clear from figure $$R = mg\cos \theta \,......\left( {\text{i}} \right)$$
Net force on the body down the inclined plane which means it is sliding downwards $$F = mg\sin \theta - f\,......\left( {{\text{ii}}} \right)$$
$$\eqalign{ & {t_2} = \sqrt {\frac{{2s}}{{g\sin \theta }}} \cr & \because {t_1} = 2{t_2}\,\,\left( {{\text{given}}} \right) \cr & \therefore t_1^2 = 4t_2^2 \cr & {\text{or}}\,\,\frac{{2s}}{{g\left( {\sin \theta - \mu \cos \theta } \right)}} = \frac{{2s \times 4}}{{g\sin \theta }} \cr & {\text{or}}\,\,\sin \theta = 4\sin \theta - 4\mu \cos \theta \cr & {\text{or}}\,\,\mu = \frac{3}{4}\tan \theta = \frac{3}{4}\tan {45^ \circ } \cr & \therefore \mu = \frac{3}{4} = 0.75 \cr} $$

$$mg\sin \theta + mg\cos \theta = ma$$

$$\eqalign{ & a = g\sin \theta + g\cos \theta = 10\left[ {\frac{3}{5} + \frac{4}{5}} \right] = 14\,m/{\sec ^2} \cr & {\text{If}}\,{f_r} = mg\sin \theta = mg \times \frac{3}{5} = \frac{{3mg}}{5} < {f_{r\max }} \cr & {f_r} < {f_{r\max }} \cr & \frac{{3mg}}{5} < \frac{{4mg}}{5} \cr} $$

Hence insect can move with constant velocity.

For smooth surface, $$s = \frac{1}{2}g\sin \theta t_1^2\,......\left( {\text{i}} \right)$$
For rough surface, $$a = g\left( {\sin \theta - \mu \cos \theta } \right)t_2^2$$
$$\therefore s = \frac{1}{2}g\left( {\sin \theta - \mu \cos \theta } \right)t_2^2\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii),
$$\eqalign{ & \therefore s = \frac{1}{2}\;g\sin \theta t_1^2 = \frac{1}{2}\;g\left( {\sin \theta - \mu \cos \theta } \right)t_2^2 \cr & {\text{Given,}}\,\,\theta = {45^ \circ }\therefore t_1^2 = \left( {1 - \mu } \right)t_2^2 \cr} $$
Also, given that, $${t_2} = n{t_1}\therefore t_1^2 = \left( {1 - \mu } \right){n^2}t_1^2$$
$$\eqalign{ & \frac{1}{{{n^2}}} = 1 - \mu \cr & \therefore \mu = \left( {1 - \frac{1}{{{n^2}}}} \right) \cr} $$

As tan $$\theta > \mu ,$$  the block has a tendency to move down the incline. Therfore a force $$P$$ is applied upwards along the incline. Here, at equilibrium $$P + f = mg\sin \theta \Rightarrow f = mg\sin \theta - P$$

Now as $$P$$ increases, $$f$$ decreases linearly with respect to $$P.$$ When $$P = mg\sin \theta ,f = 0.$$
When $$P$$ is increased further, the block has a tendency to move upwards along the incline.
Therefore the frictional force acts downwards along the incline.
Here, at equilibrium $$P = f + mg\sin \theta $$
$$\therefore f = P - mg\sin \theta $$
Now as $$P$$ increases, $$f$$ increases linearly w.r.t $$P.$$ This is represented by graph (A).

According to question, $${t_2} = n{t_1}$$
$$\eqalign{ & n\sqrt {\frac{{2d}}{{g\sin \theta }}} = \sqrt {\frac{{2d}}{{g\sin \theta - \mu g\cos \theta }}} \cr & n = \frac{1}{{\sqrt {1 - {\mu _k}} }}\,\,\,\,\,\left( {\because \cos {{45}^ \circ } = \sin {{45}^ \circ } = \frac{1}{{\sqrt 2 }}} \right) \cr & {n^2} = \frac{1}{{1 - {\mu _k}}}\,{\text{or}}\,1 - {\mu _k} = \frac{1}{{{n^2}}}\,{\text{or}}\,{\mu _k} = 1 - \frac{1}{{{n^2}}} \cr} $$

When, a cart moves with some acceleration towards right, then a pseudo force $$\left( {m\alpha } \right)$$  acts on block towards left. This force $$\left( {m\alpha } \right)$$  is action force by a block on cart. Now, block will remain static w.r.t. cart, if frictional force $$\mu R \geqslant mg$$
$$\eqalign{ & \Rightarrow \mu m\alpha \geqslant mg\,\,\left[ {{\text{as}}\,R = m\alpha } \right] \cr & \Rightarrow \alpha \geqslant \frac{g}{\mu } \cr} $$