Friction MCQ Questions & Answers in Basic Physics | Physics

$$FBD$$  of block $$A,$$

$$T - {m_1}a = {f_k}\,......\left( {\text{i}} \right)$$
$$FBD$$  of block $$B,$$

$${m_2}g - T = {m_2}a\,......\left( {{\text{ii}}} \right)$$
Adding Eqs. (i) and (i), we get
$$\eqalign{ & {m_2}g - {m_1}a = {m_2}a + {f_k} \cr & \Rightarrow {m_2}g - {m_1}a = {m_2}a + {\mu _k}{m_1}g \cr & \Rightarrow a = \frac{{\left( {{m_2} - {\mu _k}{m_1}} \right)g}}{{{m_1} + {m_2}}} \cr} $$
From Eq. (ii), $$T = {m_2}\left( {g - a} \right)$$
$$\eqalign{ & {m_2} = \left[ {1 - \frac{{\left( {{m_2} - {\mu _k}{m_1}} \right)}}{{{m_1} + {m_2}}}} \right]g \cr & T = \frac{{{m_1}{m_2}\left( {1 + {\mu _k}} \right)}}{{{m_1} + {m_2}}}g \cr} $$

Acceleration of mass at distance $$x$$
$$a = g\left( {\sin \theta - {\mu _0}x\cos \theta } \right)$$
Speed is maximum, when $$a = 0$$
$$\eqalign{ & g\left( {\sin \theta - {\mu _0}x\cos \theta } \right) = 0 \cr & x = \frac{{\tan \theta }}{{{\mu _0}}} \cr} $$

The forces acting on the block are shown. Since the block is not moving forward for the maximum force $$F$$ applied, therefore
$$F\cos {60^ \circ } = f = \mu N\,......\left( {\text{i}} \right)$$       (Horizontal Direction)
Note : For maximum force $$F,$$ the frictional force is the limiting friction $$ = \left. {\mu N} \right]$$
and $$F\sin {60^ \circ } + mg = N\,......\left( {{\text{ii}}} \right)$$
From (i) and (ii)

$$\eqalign{ & F\cos {60^ \circ } = \mu \left[ {F\sin {{60}^ \circ } + mg} \right] \cr & \Rightarrow F = \frac{{\mu mg}}{{\cos {{60}^ \circ } - \mu \sin {{60}^ \circ }}} \cr & = \frac{{\frac{1}{{2\sqrt 3 }} \times \sqrt 3 \times 10}}{{\frac{1}{2} - \frac{1}{{2\sqrt 3 }} \times \frac{{\sqrt 3 }}{2}}} = \frac{5}{{\frac{1}{4}}} = 20N \cr} $$

Limiting frictional force, $${f_l} = {\mu _s}N = 0.5 \times 5 = 2.5N.$$      But force tending to produce relative motion is the weight $$\left( W \right)$$  of the block which is less than $${f_l.}$$ Therefore, the frictional force is equal to the weight, the magnitude of the frictional force $$f$$ has to balance the weight $$0.98 N$$  acting downwards.

At limiting equilibrium, $$\mu = \tan \theta $$
$$\tan \theta = \mu = \frac{{dy}}{{dx}} = \frac{{{x^2}}}{2}\,\left( {{\text{from question}}} \right)$$

$$\eqalign{ & \because {\text{Coefficient of friction }}\mu {\text{ = 0}}{\text{.5}} \cr & \therefore 0.5 = \frac{{{x^2}}}{2} \cr & \Rightarrow x = \pm 1 \cr & {\text{Now,}}\,y = \frac{{{x^3}}}{6} = \frac{1}{6}m \cr} $$

Free body diagram of block is

From Newton's second law along $$X$$-axis
$$\eqalign{ & \sum {{F_x} = ma} \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,F - f = ma \cr & {\text{or}}\,\,F - \mu mg = ma \cr & {\text{or}}\,\,a = \frac{{F - \mu mg}}{m} \cr & {\text{Given,}}\,F = 100\,N,\mu = 0.5,\,m = 10\,kg, \cr} $$
$$g = 10\,m/{s^2}$$
Substituting, the values in the above relation for acceleration of block,
$$\eqalign{ & a = \frac{{\left( {100} \right) - \left( {0.5} \right)\left( {10} \right)\left( {10} \right)}}{{\left( {10} \right)}} \cr & = 5\,m/{s^2} \cr} $$

$$\eqalign{ & mg\sin \theta = {F_1} - \mu N \cr & N = mg\cos \theta \cr & \therefore mg\sin \theta + \mu \,mg\cos \theta = {F_1} \cr} $$
In second case (b)
$$\eqalign{ & \mu N + {F_2} = mg\sin \theta \cr & \Rightarrow \mu \,mg\cos \theta - {F_2} = mg\sin \theta \cr & {\text{or }}{F_2} = mg\sin \theta - \mu \,mg\cos \theta \,\,{\text{but}}\,\,{F_1} = 2{F_2} \cr & {\text{therefore }}mg\sin \theta + \mu \,mg\cos \theta \cr & = 2\left( {mg\sin \theta - \mu \,mg\cos \theta } \right) \cr & mg\sin \theta = 3\mu \,mg\cos \theta \cr & {\text{or}}\,\,\tan \theta = 3\mu \,\,{\text{or}}\,\,\theta = {\tan ^{ - 1}}\left( {3\mu } \right) \cr} $$

$$\eqalign{ & T = 2f + 500\,......\left( {\text{i}} \right) \cr & 8T = 5000 - f\,......\left( {{\text{ii}}} \right) \cr & f = \frac{1}{2}F\,......\left( {{\text{iii}}} \right) \cr} $$
From (i), (ii) and (iii)
$$F = \frac{{2000}}{{17}}N$$

When, static friction is present, then acceleration of body is given by $$a = - \mu g$$
Here, initial velocity $$u = 72\,km{h^{ - 1}} = 72 \times \frac{5}{{18}} = 20\,m/s$$
Final velocity $$v = 0$$
$$\therefore a = - \mu = - 0.5 \times 10 = - 5\,m/{s^2}$$
Now, from third equation of motion,
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}}\,\,{v^2} = {u^2} + 2as \cr & s = \frac{{{v^2} - {u^2}}}{{2a}} \cr & = \frac{{0 - {{\left( {20} \right)}^2}}}{{2 \times \left( { - 5} \right)}} \cr & = 40\,m \cr} $$

For the block to remain stationary with the wall
$$f = W\,\,\,\,\therefore \mu N = W$$

$$0.2 \times 10 = W \Rightarrow W = 2N$$