Laws Of Motion Practice Questions

${\text{F}}_{ {\text{k}}}=\mu _{ {\text{k}}} {\text{N}}=0.3 \times 50 \times 9.8=147 {\text{N}}$

$\; {\text{T}}_1 sin \;45^{\circ}= {\text{F}}$ $\; {\text{T}}_1 \cos 45^{\circ}= {\text{T}}_2=1 \times {\text{g}}$ $\; \therefore \tan 45^{\circ}=\;{ {\text{F}}}/{ {\text{g}}}$ $\; \therefore {\text{F}}=10 {\text{N}}$

$\;{| { {\text{F}}}^{\to }_1 |}= {\text{F}}$ $\;{| { {\text{F}}}^{\to }_{ {\text{R}}} |}={| { {\text{F}}}^{\to }_2 |}=3 {\text{F}}$ $\; {\text{F}}_{ {\text{R}}}^2= {\text{F}}_1^{2+} {\text{F}}_2^{2+}2 {\text{F}}_1 {\text{F}}_2 \cos \theta$ $\;9 {\text{F}}^2= {\text{F}}^{2+}9 {\text{F}}^{2+}6 {\text{F}}^2 \cos \theta$ $\;\cos \theta =-\;\frac{1}{6}$ $\;\theta =\cos ^{-1} (\;\frac{1}{-6})$ $\; {\text{n}}=-6$ $\;| {\text{n}}|=6$

Taking ${\text{g}}=9.8 {\text{m}} / {\ s}^2$ $\;294- {\text{N}}=30 \times 0.1$ $\; {\text{N}}=291$

${\text{a}}_{ {\text{A}}}= {\text{a}}_{ {\text{B}}}= {\text{a}}_{ {\text{C}}}=\;{ {\text{F}}}/{5+3+2}$ $=\;\frac{80}{10}=8 {\text{m}} / {\ s}^2$ ${\text{T}}_2- {\text{T}}_1=3 \times 8 \Rightarrow {\text{T}}_2=64$

${\text{f}}_{ {\text{K}}}\;=\mu {\text{mg}} \cos \theta$ $\;=0.1 \times \;{50 \times {\sqrt{3}}}/{2}$ $\;=2.5 {\sqrt{3}} {\text{N}}$ ${\text{F}}_1=\; {\text{mg}} \sin \;\theta + {\text{f}}_{ {\text{K}}}$ $=\;25+2.5 {\sqrt{3}}$ ${\text{F}}_2=\; {\text{mg}} \sin \;\theta - {\text{f}}_{ {\text{K}}}$ $\;=25-2.5 {\sqrt{3}}$ $\therefore {\text{F}}_1\;- {\text{F}}_2=5 {\sqrt{3}} {\text{N}}$

For M block $\;10 {\text{g}} sin \;53^{\circ}-\mu (10 {\text{g}}) \cos 53^{\circ}- {\text{T}}$ $=10 \times 2$ $\; {\text{T}}=80-15-20$ $\; {\text{T}}=45 {\text{N}}$ For ${\text{m}}$ block $\; {\text{T}}- {\text{mg}} sin \;37^{\circ}-\mu {\text{mg}} \cos 37^{\circ}$ $= {\text{m}} \times 2$ $\;45=10 {\text{m}}$ $\; {\text{m}}=4.5 {\text{kg}}$

${\text{F}}= {\text{ma}} \Rightarrow {\text{a}}=\;{ {\text{F}}}/{ {\text{m}}}=\;{ {\text{kt}}}/{ {\text{m}}}$a vs $t$ will be straight line passing through origin.Since option (2).

Co-efficient of friction depends on surface in contact So, depends on material of object.

$\;\;\frac{d y}{d x}=\tan \theta =\;\frac{x}{2}=\mu =\;\frac{1}{2}$ $\;x=1, y=1 / 4$

Chain block system is in equilibrium so${\text{T}}=200+100=300 {\text{N}} .$

${\text{mg}} sin \;45= {\text{f}}_{ {\text{L}}}$ ${\text{mg}} \cos 45= {\text{N}}$ ${\text{f}}_{ {\text{L}}}=\mu _{ {\ s}} {\text{N}}$ $\mu _{ {\ s}}=\tan 45=1$or$\tan \theta =\mu _{ {\ s}}$ ( $\theta$ is angle of repose)$\tan 45=\mu _{ {\ s}}=1$correct option (1)

$R=$ net reaction force by shoulderBalancing torque about pt of contact on ground:$\; {\text{W}} (\;{ {\text{L}}}/{2} \cos \theta ) = {\text{R}}( {\text{L}} \cos \theta )$ $\;\Rightarrow {\text{R}}=\;{ {\text{W}}}/{2}$

Case-1 : No friction$\; {\text{a}}= {\text{g}} sin \;\theta$ $\;ℓ=\;\frac{1}{2}( {\text{g}} sin \;\theta ) {\text{t}}_1^2$ $\; {\text{t}}_1=\sqrt{\;{2 ℓ}/{ {\text{g}} sin \;\theta }}$ Case-2 : With friction$\; {\text{a}}= {\text{g}} sin \;\theta -\mu {\text{g}} \cos \theta$ $\;ℓ=\;\frac{1}{2}( {\text{g}} sin \;\theta -\mu {\text{g}} \cos \theta ) {\text{t}}_2^2$ $\;\sqrt{\;{2 ℓ}/{ {\text{g}} sin \;\theta -\mu {\text{g}} \cos \theta }}= {\text{n}} \sqrt{\;{2 ℓ}/{ {\text{g}} sin \;\theta }}$ $\;\mu =1-\;{1}/{ {\text{n}}^2}$

$\;a_{\;\text{sys }\;}= (\;\frac{m_2-m_1}{m_1+m_2}) g=\;\frac{g}{8}$ $\;\Rightarrow \;\frac{m_2}{m_1}=\;\frac{9}{7}$

\;{ {\text{P}}}^{\to }=\cos ( {\text{kt}}) { {\text{i}}^{\^}}- sin \;( {\text{kt}}) { {\text{j}}^{\^}} ;|{ {\text{P}}}^{\to }|=1 $\; \because { {\text{P}}}^{\to }= {\text{m}} { {\text{v}}}^{\to }$ \; \therefore { {\text{P}}^{\^}}={ {\text{v}}^{\^}} \;\Rightarrow { {\text{v}}^{\^}}=\cos ( {\text{kt}}) { {\text{i}}^{\^}}- sin \;( {\text{kt}}) { {\text{j}}^{\^}} \;{ {\text{a}}^{\^}}=\;{- {\text{k}} sin \;( {\text{kt}}) { {\text{i}}^{\^}}- {\text{k}} \cos ( {\text{kt}}) { {\text{j}}^{\^}}}/{ {\text{k}}} \;\Rightarrow { {\text{a}}^{\^}}=- sin \;k t { {\text{i}}^{\^}}-\cos {\text{kt}} { {\text{j}}^{\^}} \; \because { {\text{F}}^{\^}}={ {\text{a}}^{\^}}=- sin \; {\text{kt}} { {\text{i}}^{\^}}-\cos {\text{kt}} { {\text{j}}^{\^}} \;\cos \theta =\;{{ {\text{F}}^{\^}} \cdot { {\text{P}}^{\^}}}/{|{ {\text{F}}^{\^}}| { {\text{P}}^{\^}} \mid }=-\;{ sin \;k t \cos {\text{t}}+ sin \; {\text{kt}} \cos {\text{t}}}/{1 \times 1}=0 $\;\Rightarrow \theta =\;\frac{\pi }{2}$

To find the normal reaction ($N$), we analyze the vertical components of the forces acting on the garden roller.

1. Identify the Forces:

• Weight ($W$): Acts vertically downwards.
  $W = mg = 70 \times 10 = 700\text{ N}$
• Vertical Component of Push ($F_y$): Since the roller is being pushed at an angle of $30^\circ$ to the horizontal, the downward vertical component is:
  $F_y = F \sin(30^\circ) = 200 \times \frac{1}{2} = 100\text{ N}$
• Normal Reaction ($N$): Acts vertically upwards from the ground.

2. Vertical Equilibrium Equation:
The sum of upward forces must equal the sum of downward forces:
$N = W + F_y$
$N = mg + F \sin(30^\circ)$

3. Final Calculation:
$N = 700 + 100 = 800\text{ N}$

The normal reaction on the roller is 800 N.

Correct Option: (C)

Statement-1When elevator is moving with uniform speed ${\text{T}}= {\text{F}}_{ {\text{g}}}$Statement-2When elevator is going down with increasing speed, its acceleration is downward.Hence ${\text{W}}- {\text{N}}=\;{ {\text{W}}}/{ {\text{g}}} \times {\text{a}}$ ${\text{N}}= {\text{W}} (1-\;{ {\text{a}}}/{ {\text{g}}})$ i.e. less than weight.