Physics M.C.Q (1 Marks)

$\overrightarrow{ V }_{ F }=( V ) \text { Eastward }$

$\overrightarrow{\Delta V }=\overrightarrow{ V }_{ F }-\overrightarrow{ V }_{ i }$

$=\text { Along North - East }$

$f _{ L }= ma a _{\max }$

$\mu mg = ma _{\max }$

$a _{\text {max }}=\mu g$

$=0.15(10)$

$=1.5\,m / s ^2$

$T - f = m _{ T } a$

$f=\mu m_{T} g$

$f=0.05 \times 10 \times 10$

$f=5 N$

$T -5=10 a$

F.B.D. of block

$m_{b} g-T=m_{b} a$

$2 \times 10-T=2 a$

$20- T =2 a$

Equation (i) $+$ (ii)

$15=12 a$

$a =\frac{15}{12} \Rightarrow a =1.25 m / s ^{2}$

Resultant $=\sqrt{\mathrm{N}^{2} +\mathrm{f}^{2}} =\sqrt{(\mathrm{mg})^{2}+\mathrm{f}^{2}} \leq \mathrm{mg} \sqrt{1+\mu^{2}}$

$\mathrm{f}_{\mathrm{s}}=\mathrm{mg}$

As $\mathrm{f}_{\mathrm{s}} \leq \mathrm{f}_{\mathrm{L}}$

$\Rightarrow \mathrm{mg} \leq \mu \mathrm{mr} \mathrm{ \omega}^{2}$

$\Rightarrow \omega \geq \sqrt{\frac{g}{\mu r}}$

$\Rightarrow \omega_{\min }=10\; \mathrm{rad} / \mathrm{s}$

$f=\mu_s N $

$\mu_s=\frac{f}{N}$

$v_{m}=\sqrt{\mu r g}=\sqrt{0.2 \times 3 \times 10}=2.45 m / s$

$=\frac{2.45 \times 60 \times 60}{1000} km / h =8.82 km \cdot h^{-1}$

Among the given options $7.2\; km \cdot h^{-1}$ is less than this.

$\begin{array}{l}
 \Rightarrow \,\frac{{{v^2}_{\max }}}{{Rg}} = \frac{{N\sin \theta  + {\mu _s}N\cos \theta }}{{N\cos \theta  - {\mu _s}N\sin \theta }}\\
\,\,\,\,\,\,{v_{\max }} = \sqrt {Rg\left( {\frac{{{\mu _s} + \tan \theta }}{{1 - {\mu _s}\tan \theta }}} \right)} 
\end{array}$

$\Rightarrow \mathrm{a}=20 \mathrm{m} / \mathrm{s}^{2}$

$f_{max}=0.1\times 1000=100\,N$

$mg=50\,N$

$f=50\,N$

$\mathrm{f}_{\mathrm{B}}=100 \mathrm{N}+20 \mathrm{N}=120 \mathrm{N}$

i.e., frictional force applied by the wall on the block $B$ is $120 N$

$=10\left[\frac{1}{2}-(0.2) \frac{\sqrt{3}}{2}\right] 5=16.34 \mathrm{ms}^{-1} $

$F(t=2)=8 N$

as $\mathrm{F}<\mathrm{F}_{\max } \Rightarrow$ friction $=8 \mathrm{N}$

$F=f+\mathrm{ma}$

Power $(\mathrm{P})=\mathrm{Fv}=(\mathrm{f}+\mathrm{ma}) \mathrm{v}=(\mu \mathrm{mg}+\mathrm{ma})$ at

$t'=nt$   $\mathrm{n}=1 \frac{1}{3}=\frac{4}{3}$

$\mu=\tan 45\left[1-\frac{1}{(4 / 3)^{2}}\right]$

$\mu=\frac{7}{16}$

In equilibrium of block $B$ implies $T=M g \ldots(i)$

If block $A$ does not move, then $T=f_{s}$

where $f_{s}=$ frictional force $=\mu_{S} N=\mu_{s} m g$

implies $T=\mu_{s} m g \ldots(i i)$

Thus, from Eqs. $( i )$ and $(ii),$ we have

$M=\mu_{s} m g$ or $M=\mu_{s} m$

Given: $\mu_{s}=0.2, m=2 k g$

$\therefore M=0.2 \times 2=0.4 k g$

$\therefore \mu_{s}=\frac{1}{3}=0.33$

$\therefore a=\frac{\mathrm{f}}{\mathrm{m}}=\frac{\mu \mathrm{mg}}{\mathrm{m}}=\mu \mathrm{g}=0.5 \times 10=5 \mathrm{ms}^{-2}$

Using, $\mathrm{v}^{2}-\mathrm{u}^{2}=2 \mathrm{aS}$

${0^{2}-2^{2}=2(-5) \times 5} $

${\mathrm{S}=0.4 \mathrm{m}}$

${F=\mu R=0.5 \times m g=0.5 \times 60=30 \mathrm{N}}$

$\mathrm{Now}$ $\mathrm{F}=\mathrm{T}_{1}=\mathrm{T}_{2} \cos 45^{\circ}$

$\text { or } \quad 30=\mathrm{T}_{2} \cos 45^{\circ}$

${\text { and }}  {W=T_{2} \sin 45^{\circ}} $

${\therefore \quad}  {W=30 \mathrm{N}}$

$\left(m_1+m_2\right) a_{\text {max }}=f$

$\rightarrow F=\mu N$

$N=m_1 g$

$m_2 a_{\text {max }}=\mu m_1 g$

$a_{\text {max }}=\mu \frac{m_1}{m_2} g$

$F=\left(m_1+m_2\right) \mu \frac{m_1}{m_2} g$

$= m_2\left(\frac{m_1}{m_2}+1\right) \mu \frac{m_1}{m_2} g$

$F =\mu m_1\left(\frac{m_1}{m_2}+1\right)$

$N+F_2=m g$

$N=m g-F_2$

As $F_2$ increases $N$ will decrease

Static friction $f_{ s }=\mu_s N=\mu_s\left(m g-F_2\right)$

$\Rightarrow$ By increasing $F_2, f_s$ will decrease hence the block will slide

Limiting friction $F_L=(0.3)(1)(g)$

$=3 \,N$

$x$-component or horizontal component of force is $=1 \,N$

hence this much of magnitude will act in backward direction due to friction.

If both move together $a=\frac{10}{101} \simeq 0.1 \,m / s ^2$

Now, $F_{\text {net }}=1(0.1)=0.1 \,N$

$f_L=(0.5)(1)(g)=5 \,N$

So, $f=0.1 \,N$

$ma = F - {\mu _k}mg$

${\mu _k} = \frac{{F - ma}}{{mg}} = \frac{{129.4 - 10 \times 10}}{{10 \times 9.8}} = 0.3$

$ = ma + \mu \;mg = m(a + \mu \;g) = (1500 + 500)(1 + 0.2 \times 10)$ 

$ = 2000 \times 3 = 6000\;N$

so mass of the body $m = 6.4\;kg$,   ${\mu _s} = 0.6$, ${\mu _k} = 0.4$

Net acceleration $ = \frac{{{\rm{Applied\, force - Kinetic\, friction}}}}{{{\rm{Mass\, of\, the \,body}}}}$

$ = \frac{{{\mu _s}mg - {\mu _k}mg}}{m} = ({\mu _s} - {\mu _k})g = (0.6 - 0.4)g = 0.2\,g$

$a = \frac{{F - \mu mg}}{m} = \frac{{129.4 - 0.3 \times 10 \times 9.8}}{{10}} = 10\;m/{s^2}$

$t = \frac{{u \times m}}{F}$$ = \frac{{30 \times 1000}}{{5000}} = 6\;\sec $

If body is moving on rough surface

Frictional force $=\left( f _{ s }\right)_{\max }= f _{ k }$

$=\mu mg$

$=0.6 \times 10 \times 10$

$=60\,N$

Since friction is self adjusting force maximum forc eof $60\,N$ is acting, but here only External force $=20\,N$ is acting so friction can acts is $20\,N$

If external force increases then $f$ will increase and go upto $60\,N$ beyond which object will move

$a=2\,m / s ^2$

$g=9.8\,m / s ^2$

$Friction force = \mu g$

$=\mu \times 10 \times 9.8$

$=98\,\mu$

Net force $=$ External force $=$ friction force

$m \times a =40-98\,\mu$

$2 \times 10 =40-98\,\mu$

$98 \mu =40-20$

$=20$

$\mu=\frac{20}{98}=\frac{10}{49}=0.20$

It both are moving together

$a=\frac{F}{8}$

for $3 \,kg$ block

$f=3\left(\frac{F}{8}\right)$

$(0.5 (3) g=\frac{3 F}{8}$

$F=40 \,N$

So, $m=4 \,kg$

$N=F \sin 30^{\circ}=F / 2$

$F \cos 30^{\circ}+\mu N=(10) g$

$\frac{F \sqrt{3}}{2}+0.5\left(\frac{F}{2}\right)=100$

$F\left(\frac{2 \sqrt{3}+1}{4}\right)=100$

$F \simeq 89.7 \,N$

Limiting friction = ${F_l} = {\mu _s}R = {\mu _s}mg\cos \theta $

$= 0.4 × 102 × 9.8 × cos\,30 = 346\, N$

As net external force is less than limiting friction therefore friction on the body will be $250 N.$

Normal reaction $ = mg\cos \theta $ 

Given : $mg\cos \theta = 2mg(\sin \theta - \mu \cos \theta )$ 

By solving $\theta = {45^o}$.

$ = 10 \times 9.8(\sin 30^\circ + 0.5\cos 30^\circ ) = 91.4\;N$.

Let it travel distance ‘S’ before coming to rest

$S = \frac{{{v^2}}}{{2\mu g}} = \frac{{20}}{{2 \times 0.25 \times 10}} = 4\,m$

act at $A.$

about $G$

$\mathrm{N} \times \frac{\mathrm{a}}{2}=\mu \mathrm{N} \times \frac{\mathrm{a}}{2}$

$\mu=1$

$\theta=45^{\circ}$        $...(i)$

and $\operatorname{mg} \sin \theta \leq f_{\mathrm{r} \max }=\mu \mathrm{mg} \cos \theta$

$\Rightarrow \mu \geq \tan \theta$

$\Rightarrow \mu \geq 1$

Downward force $=m g \sin 30=1 \times 9.8 \times 0.5=4.9 \mathrm{N}$

since downward force is less than maximum frictional force, thus the tension in the string is zero.

$\therefore \quad\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g} \sin \theta-\left(\mu \mathrm{m}_{1} \mathrm{g} \cos \theta+\mu \mathrm{m}_{2} \underline{\mathrm{g}} \cos \theta\right)$

$=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{a}_{\mathrm{net}}$

$\therefore a_{n e t}=g\left(\sin \theta-\frac{\left(\mu m_{1}+\mu m_{2}\right) \cos \theta}{m_{1}+m_{2}}\right)$

Here, $m_{1}=4 \mathrm{kg}, \mathrm{m}_{2}=2 \mathrm{kg}, \theta=30^{\circ} ; \mu=0.3$

$\mu_{s}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=0.6$

$a=g \sin 30^{\circ}-\mu_{k} g \cos 30^{\circ}$

$S=u t+\frac{1}{2} a t^{2}$

$\Rightarrow 4=\frac{1}{2}\left[\frac{\mathrm{g}}{2}-\frac{\mu_{\mathrm{k}} \mathrm{g} \sqrt{3}}{2}\right] \times 16 \Rightarrow \mu_{\mathrm{k}}=0.5$

$=g\left[\sin 45^{\circ}-0.5 \times \cos 45^{\circ}\right]$

$=\frac{4.9}{\sqrt{2}} \mathrm{m} / \mathrm{s}^{2} 1 $

$\sum F y=0$

$\Rightarrow N - mg \cos \theta=0$

$\Rightarrow N = mg \cos \theta----(1)$

$Fx =0$ (as constant velocity $\Rightarrow$ No acceleration)

$\Rightarrow mg \sin \theta-\mu_{ k } N =0$

$\Rightarrow mg \sin \theta-\mu_{ k } mg \cos \theta=0$

$\Rightarrow \cos \theta\left(\tan \theta-\mu_{ k }\right)=0$

$\Rightarrow \mu_{ k }=\tan \theta=\tan 37^{\circ}=\frac{3}{4}=0.75$