Laws Of Motion Practice Questions

$\; {\text{N}}= {\text{Mg}}- {\text{F}} {\text{Sin}} 30^{\circ}$ $\;= {\text{mg}}-\;{ {\text{F}}}/{2}=100-\;{ {\text{F}}}/{2}=\;{200- {\text{F}}}/{2}$ $\; {\text{F}} {\text{Cos}} 30^{\circ}=\mu {\text{N}}$ $\;{\sqrt{3}} \;{ {\text{F}}}/{2}=0.25 \times (\;{200- {\text{F}}}/{2})$ $\;4 {\sqrt{3}} {\text{F}}=200- {\text{F}}$ $\; {\text{F}}=\;{200}/{4 {\sqrt{3}}+1}=25.22$

$F_1\;=10 {\text{N}}$ $F_2\;=8 {\text{N}}$ $F_3\;=6 {\text{N}}$ { {\text{a}}}^{\to }\;=\;{ {\text{F}}_2 { {\text{i}}^{\^}}+ {\text{F}}_3 { {\text{j}}^{\^}}}/{5} { {\text{a}}}^{\to }\;=\;{8 { {\text{i}}^{\^}}+6 { {\text{j}}^{\^}}}/{5} $|{ {\text{a}}}^{\to }|\;=\;{{\sqrt{64+36}}}/{5}$ $\;=\;\frac{10}{5}=2 {\text{m}} / {\ sec}^2$OR ${\text{F}}_1^2= {\text{F}}_2^{2+} {\text{F}}_3^2$Now ${\text{F}} 1$ is removed then net force$\; {\text{F}}_{\;\text{net }\;}=\sqrt{ {\text{F}}_2^{2+} {\text{F}}_3^2}$ $\; \;\; ={\sqrt{64+36}}$

$\;\theta =30^{\circ}$ $\;\cos \theta =\;{{\sqrt{3}} {\text{g}}}/{ {\text{T}}}$ $\;\Rightarrow \;{{\sqrt{3}}}/{2}=\;{{\sqrt{3}} {\text{g}}}/{ {\text{T}}}$ $\;\Rightarrow {\text{T}}=20 {\text{N}}$

${\text{Mg}} sin \;30^{\circ}-\mu {\text{mg}} \cos 30^{\circ}= {\text{ma}}$ $\;\frac{g}{2}-\;{{\sqrt{3}}}/{2} \cdot \mu g=\;\frac{g}{4}$ $\;\;{{\sqrt{3}}}/{2} \mu =\;\frac{1}{4}$ $\;\mu =\;{1}/{2 {\sqrt{3}}}$

$4 g sin \;60^{\circ}-T=4 a$ . . . (1) $4 {\text{g}} sin \;60^{\circ}$ ${\text{T}}- {\text{g}} sin \;30^{\circ}= {\text{a}}$ . . . (2)Solving (1) and (2) we get.$\;20 {\sqrt{3}}-T=4 T-20$ $\;T=4({\sqrt{3}}+1) N$

Impulse $=$ Area under $F=t$ curve(a) $\;\frac{1}{2} \times 1 \times 0.5=\;\frac{1}{4} {\text{N}} . {\ s}$(b) $0.5 \times 2=1$ N.s (maximum)(c) $\;\frac{1}{2} \times 1 \times 0.75=\;\frac{3}{8} {\text{N}} . {\ s}$(d) $\;\frac{1}{2} \times 2 \times 0.5=\;\frac{1}{2}$ N.s

$\; {\text{a}}_1= {\text{g}} sin \;\theta = {\text{g}} / {\sqrt{2}}$ $\; {\text{a}}_2= {\text{g}} sin \;\theta - {\text{Kg}} \cos \theta =\;{ {\text{g}}}/{{\sqrt{2}}}-\;{ {\text{Kg}}}/{{\sqrt{2}}}$ \; {\text{t}}_2= {\text{nt}}_1 & \;\; {\text{a}}_1 {\text{t}}_1^2= {\text{a}}_2 {\text{t}}_2^2 $\;\;{ {\text{g}}}/{{\sqrt{2}}} {\text{t}}_1^2= (\;{ {\text{g}}}/{{\sqrt{2}}}-\;{ {\text{kg}}}/{{\sqrt{2}}}) {\text{n}}^2 {\text{t}}_1^2$ $\; {\text{K}}=1-\;{1}/{ {\text{n}}^2} \;\; \;\text{ Ans. }\; 4$

$\; {\text{F}}_1= {\text{mg}} sin \;45^{\circ}+ {\text{f}}= {\text{mg}} sin \;45^{\circ}+\mu {\text{N}}$ $\; {\text{F}}_1=\;{ {\text{mg}}}/{{\sqrt{2}}}+\mu {\text{mg}} \cos 45^{\circ}$ $\; {\text{F}}_1=\;{ {\text{mg}}}/{{\sqrt{2}}}(1+\mu )$ $\;F_2=m g sin \;45^{\circ}- {\text{f}}= {\text{mg}} sin \;45^{\circ}-\mu {\text{N}}$ $\;=\;{ {\text{mg}}}/{{\sqrt{2}}}(1-\mu )$ $\; {\text{F}}_1=2 {\text{F}}_2$ $\;\;{ {\text{mg}}}/{{\sqrt{2}}}(1+\mu )=2 \;{ {\text{mg}}}/{{\sqrt{2}}}(1-\mu )$ $\;1+\mu =2-2 \mu$ $\;\mu =1 / 3=0.33$

$\;50=V \times 10$ $\;V=5 {\text{m}} / {\ s}$ $\; {\text{V}}=0+ {\text{a}} \times 20$ $\;5= {\text{a}} \times 20$ $\; {\text{a}}=\;\frac{1}{4} {\text{m}} / {\ s}^2$ $\; {\text{F}}= {\text{ma}}=20 \times \;\frac{1}{4}=5 {\text{N}}$

$\; {\text{v}}=10 \sqrt{ {\text{x}}} \Rightarrow \;{ {\text{dv}}}/{ {\text{dx}}}=\;{10}/{2 \sqrt{ {\text{x}}}}=\;{5}/{\sqrt{ {\text{x}}}}$ $\; {\text{a}}= {\text{v}} \;{ {\text{dv}}}/{ {\text{dx}}}$ $\; {\text{a}}= {\text{v}} \times \;{5}/{\sqrt{ {\text{x}}}}=10 \sqrt{ {\text{x}}} \times \;{5}/{\sqrt{ {\text{x}}}}=50 {\text{ms}}^{-2}$ $\; {\text{F}}= {\text{ma}}=\;\frac{500}{1000} \times 50=25 {\text{N}}$

Given, $\vec{r}=10 t \hat{i}+15 t^2 \hat{j}+7 \hat{k}$ \;{ {\text{v}}}^{\to }=\;\frac{\vec{d r}}{d t}=10 \hat{j}+30 { {\text{j}}^{\^}} \;{ {\text{a}}}^{\to }=\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}}=30 { {\text{j}}^{\^}} $\;{ {\text{F}}}^{\to }= {\text{m}} { {\text{a}}}^{\to } \to \;\text{ along }\;(+) {\text{y}} \;\text{-axis }\;$

{ {\text{V}}}^{\to }= (2 t \hat{i}+3 {\text{t}}^2 { {\text{j}}^{\^}}) {\text{m}} / {\ s}, \;\text{ mass }\; {\text{m}}=500 {\text{gm}} ${\;\text{ Force, }\;}^{\to } { {\text{F}}}^{\to }= {\text{ma}}$ -\;{ {\text{F}}}^{\to }=\;\frac{1}{2} (\;{ {\text{d}} { {\text{v}}}^{\to }}/{ {\text{dt}}}) \Rightarrow { {\text{F}}}^{\to }=\;\frac{1}{2}(2 { {\text{i}}^{\^}}+6 { {\text{t}}^{\^}}) \;{ {\text{F}}}^{\to }=({ {\text{i}}^{\^}}+3 { {\text{t}}^{\^}}) \;\;\text{ At }\; t=1 {\ s} \Rightarrow { {\text{F}}}^{\to }=({ {\text{i}}^{\^}}+3 { {\text{j}}^{\^}}) $\; {\text{x}}=3$

${\text{a}}= {\text{g}} \ sin \theta$ $ℓ=\;\frac{1}{2} {\text{g}} \ sin 30^{\circ}(2)^2$ $ℓ=\;\frac{1}{2} {\text{g}} \ sin 45^{\circ} {\text{t}}^2$ $(\;\frac{1}{2})(4)=\;\frac{1}{\sqrt{2}} {\text{t}}^2 \Rightarrow {\text{t}}=\sqrt{2 \sqrt{2}} ≃ 1.68$

$\therefore a_{\max }=\mu {\text{g}}$ $=0.5 \times 9.8=4.9 {\text{m}} / {\ s}^2$ $\therefore {\text{F}}_{\max }=(8+2) \times 4.9=49 {\text{N}}$

$a=\;\frac{m_2-m_1}{m_1+m} g=\;\frac{g}{2}$ $\Rightarrow m_2=3 m_1$ $(L-l)=3 l$ $L=4 l$ $l=\;\frac{L}{4}$

Case 1 ${\text{a}}=\;{ {\text{m}}_2 {\text{g}}- {\text{m}}_1 {\text{g}}}/{ {\text{m}}_1+ {\text{m}}_2}$ ${\text{M}}_2=2 {\text{m}}_1$ $a_1=\;\frac{2 m_1 g-m_1 g}{3 m_1}$ ${\text{a}}_1= {\text{g}} / 3$ Case $-2$ ${\text{M}}_2=3 {\text{m}}_1$ ${\text{a}}_2=\;{3 {\text{m}}_1 {\text{g}}- {\text{m}}_1 {\text{g}}}/{4 {\text{m}}_1}$ ${\text{a}}_2=\;{ {\text{g}}}/{2}$ $\;{ {\text{a}}_1}/{ {\text{a}}_2}=\;{\;{ {\text{g}}}/{3}}/{{ {\text{g}}}/{2}}=\;\frac{2}{3}$

Here $F_{\max }$ force is trying to move the block upward so friction force will be applied towards downward.Along horizontal direction,$N=50 N$Along vertical diretion,$F_{\max }=m g+f_{\max }$ $=2 \times 10+\mu {\text{N}}$ $=20+0.5 \times 50$ $=20+25$ $=45 {\text{N}}$

${\text{N}}_1= {\text{mg}}$ ${\text{N}}_2= {\text{mg}}+ {\text{ma}}$ ${\text{N}}_3= {\text{mg}}- {\text{ma}}$ ${\text{N}}_4= {\text{mg}}$ ${\text{N}}_5= {\text{mg}}$