Laws Of Motion Practice Questions

When the block moves up the inclined plane, the component of gravity acting downwards along the plane is $mg \sin \theta$. Since this force acts opposite to the direction of motion, the retardation is $a = g \sin \theta$. Using the kinematic equation $v^2 = u^2 + 2aS$, where $v = 0$ at maximum distance $S$:
$0^2 = u^2 - 2(g \sin \theta)S$
$u^2 = 2g \sin \theta S$
$S = \frac{u^2}{2 g \sin \theta}$

The forces acting on the particle are gravity $mg$ downwards and drag $kv$ upwards. Applying Newton's second law: $m \frac{dv}{dt} = mg - kv$.
Separating variables and integrating with initial condition $v(0)=0$:
$\int_{0}^{v} \frac{dv}{mg - kv} = \int_{0}^{t} \frac{dt}{m} \implies -\frac{1}{k} \ln\left(\frac{mg - kv}{mg}\right) = \frac{t}{m}$
Solving for velocity gives: $v(t) = \frac{mg}{k}(1 - e^{-\frac{k}{m}t})$.
As $t \to 0$, $v \to 0$, and as $t \to \infty$, $v \to \frac{mg}{k}$ (terminal velocity).
The graph starts at the origin, increases with a decreasing slope, and asymptotically approaches the horizontal line $v = \frac{mg}{k}$, matching option D.

To find the force $\vec{F}$ required to slide block $C$, we calculate the normal and frictional forces acting on the system:

1. Normal Forces:

   • Between $A$ and $B$: $N_1 = m_A g = 4 \times 10 = 40\text{ N}$.
   • Between $B$ and $C$: $N_2 = (m_A + m_B) g = (4 + 6) \times 10 = 100\text{ N}$.
   • Between $C$ and ground: $N_3 = (m_A + m_B + m_C) g = (4 + 6 + 8) \times 10 = 180\text{ N}$.

2. Frictional Forces ($\mu = 0.5$):

   • $f_1 = \mu N_1 = 0.5 \times 40 = 20\text{ N}$.
   • $f_2 = \mu N_2 = 0.5 \times 100 = 50\text{ N}$.
   • $f_3 = \mu N_3 = 0.5 \times 180 = 90\text{ N}$.

3. Total Force Calculation:
   Since block $B$ is connected to the wall, the tension $T$ in the rope balances the friction forces on $B$ (from $A$ and $C$), so $T = f_1 + f_2 = 20 + 50 = 70\text{ N}$. The force $\vec{F}$ required to slide block $C$ must overcome the friction from the ground ($f_3$), the friction from block $B$ ($f_2$), and the tension transferred through the system ($T$):
   $F = f_3 + f_2 + T = 90 + 50 + 70 = 210\text{ N}$

[CORRECT_OPTION: N/A]

Considering the equilibrium of one half of the chain (mass $\frac{m}{2}$), the vertical forces must balance: $T_{support} \sin(30^\circ) = \frac{mg}{2}$, where $T_{support}$ is the tension at the support.
The horizontal forces must also balance, giving the tension at the lowest point as $T_0 = T_{support} \cos(30^\circ)$.
From the vertical equilibrium, $T_{support} = \frac{mg/2}{\sin(30^\circ)} = \frac{mg/2}{1/2} = mg$.
Substituting this into the horizontal equilibrium equation yields $T_0 = mg \cos(30^\circ) = mg \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2} mg$.

In the non-inertial reference frame of the inclined plane, a pseudo-force $F_p = m a_0$ acts on the block towards the right. The components of acceleration along the incline (downwards) are $g \sin \theta$ due to gravity and $-a_0 \cos \theta$ due to the pseudo-force. Thus, the net acceleration is $a = g \sin \theta - a_0 \cos \theta$.

The length of the inclined surface is $s = \frac{L}{\cos \theta}$. Using the kinematic equation $s = \frac{1}{2} a t^2$, we have:
$t = \sqrt{\frac{2s}{a}} = \sqrt{\frac{2(L/\cos \theta)}{g \sin \theta - a_0 \cos \theta}} = \sqrt{\frac{2L}{g \sin \theta \cos \theta - a_0 \cos^2 \theta}}$

To match the given options, multiply the numerator and denominator by 2:
$t = \sqrt{\frac{4L}{2g \sin \theta \cos \theta - 2a_0 \cos^2 \theta}}$

Using the trigonometric identities $2 \sin \theta \cos \theta = \sin 2\theta$ and $2 \cos^2 \theta = 1 + \cos 2\theta$:
$t = \sqrt{\frac{4L}{g \sin 2\theta - a_0(1 + \cos 2\theta)}}$

Given the mass $m = 4 \text{ kg}$ and force $\vec{F} = (4t^3 \hat{i} - 3t \hat{j}) \text{ N}$, the acceleration is $\vec{a} = \frac{\vec{F}}{m} = (t^3 \hat{i} - \frac{3}{4}t \hat{j}) \text{ m/s}^2$.
Integrating acceleration with respect to time, $\vec{v}(t) = \int \vec{a} \, dt = (\frac{t^4}{4} \hat{i} - \frac{3t^2}{8} \hat{j}) \text{ m/s}$, as $\vec{v}(0) = 0$.
At $t = 2 \text{ s}$, the velocity is $\vec{v}(2) = \frac{2^4}{4} \hat{i} - \frac{3(2^2)}{8} \hat{j} = 4 \hat{i} - \frac{12}{8} \hat{j} = 4 \hat{i} - \frac{3}{2} \hat{j} \text{ m/s}$.
Integrating velocity, $\vec{r}(t) = \int \vec{v} \, dt = (\frac{t^5}{20} \hat{i} - \frac{t^3}{8} \hat{j}) \text{ m}$, as $\vec{r}(0) = 0$.
At $t = 2 \text{ s}$, the position is $\vec{r}(2) = \frac{2^5}{20} \hat{i} - \frac{2^3}{8} \hat{j} = \frac{32}{20} \hat{i} - \frac{8}{8} \hat{j} = \frac{8}{5} \hat{i} - \hat{j} \text{ m}$.

The equilibrium conditions for the mass $m$ are given by resolving forces horizontally ($T_1 \cos \theta_1 = T_2 \cos \theta_2$) and vertically ($T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg$). Given $T_1 = \sqrt{3} T_2$, the horizontal equation becomes $\sqrt{3} \cos \theta_1 = \cos \theta_2$. Testing Option C where $\theta_1 = 60^{\circ}$ and $\theta_2 = 30^{\circ}$, we find $\sqrt{3} \cos 60^{\circ} = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} = \cos 30^{\circ}$, which is satisfied. Substituting these into the vertical equation gives $T_2 (\sqrt{3} \sin 60^{\circ} + \sin 30^{\circ}) = mg$, which simplifies to $T_2 (\sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2}) = T_2 (\frac{3}{2} + \frac{1}{2}) = 2T_2 = mg$. Thus, $T_2 = \frac{mg}{2}$.

$\; {\text{W}}=\Delta k \;\; \;\text{ so }\;{| {\text{W}}_f |}=\;\text{ loss \in }\; {\text{KE}}$ $\;\;\frac{1}{2} k ((1.9)^{2-}(0.1)^2) =\;\frac{1}{2} \times 1 \{10^{2-}v^2\}$ $\;10 \times 1.8 \times 2=100-v^2$ $\;v^2=64$ $\;v=8 {\text{m}} / {\ s}$

$\;\vec{F} \cdot t=m (\vec{v}_2-\vec{v}_1)$ $\;\;\frac{6}{2} \times \;\frac{5}{3} {k}^{∧}=\vec{v}_2-\vec{v}_1$ $\;\vec{v}_2=\vec{v}_1+5 {k}^{∧}=3 {i}^{∧}+4 {j}^{∧}+5 {k}^{∧}$

$\;a=\;\frac{m g sin \;\theta -\mu m g \cos \theta }{m}=\;{g {\sqrt{3}}}/{2}-\;\frac{\mu g}{2}$ $\;\Rightarrow \;\frac{g}{2}=\;\frac{g}{2}({\sqrt{3}}-\mu ) \Rightarrow {\sqrt{3}}-\mu =1$ $\;\Rightarrow \mu =({\sqrt{3}}-1) \approx 1.73-1=0.73$

$\;\;\frac{d x}{d t}=4 {\sqrt{x}}$ $\;x=4 t^2$ $\;v=\;\frac{d x}{d t}=8 t$ $\;a=8 {\text{m}} / {\ s}^2$ $\;F=m a=4 {\text{N}}$

Let us assure buoyancy force on balloon be $B$.$\;B-M g=M a$ $\;B-(M-m) g=(M-m) 3 a$Subtracting (ii) from (i)$\;(B-M g)-(B-(M-m) g)=M a-3(M-m) a$ $\;-m g=(3 m-2 M) a$or $m=\;\frac{2 M a}{3 a+g}$

$\;T_1 sin \;60+T_2 sin \;30=m g$ $\;T_1 \cos 60=T_2 \cos 30$ $\;T_1={\sqrt{3}} T_2$ $\;{\sqrt{3}} T_2 \times \;{{\sqrt{3}}}/{2}-\;\frac{T_2}{2}=m g$ $\;T_2=\;\frac{m g}{2}=5 {\text{N}}$ $\;T_1=\;{{\sqrt{3}} m g}/{2}=5 {\sqrt{3}} {\text{N}}$

1. Identify the Forces

• Pulling Force (Weight of block): $m_B \times g = 6 \times 10 = 60\text{ N}$
• Opposing Force (Kinetic Friction): $f_k = \mu_k \times N$
  Since the trolley is on a horizontal surface, $N = m_T \times g = 20 \times 10 = 200\text{ N}$.
  $f_k = 0.04 \times 200 = 8\text{ N}$

2. Calculate Acceleration ($a$)

Using Newton's Second Law for the entire system (Trolley + Block):
$a = \frac{F_{\text{net}}}{\text{Total Mass}} = \frac{W_B - f_k}{m_T + m_B}$

Substituting the values:
$a = \frac{60 - 8}{20 + 6} = \frac{52}{26} = 2\text{ m/s}^2$

The acceleration of the system is 2 m/s$^2$.

Correct Option: (C)

$\;\;\text{ Net force }\;=8 \hat{i}+4 \hat{j}+4 \hat{k}$ $\;\vec{a}=\;\frac{\vec{F}}{m}=2 \hat{i}+\hat{j}+\hat{k}$

$\;a=\;\frac{ (m_1-m_2) g}{ (m_1+m_2) }=\;\frac{g}{8}$ $\;8 m_1-8 m_2=m_1+m_2$ $\;7 m_1=9 m_2$ $\;\;\frac{m_1}{m_2}=\;\frac{9}{7}$

$\;a= (\;\frac{M_2-M_1}{M_1+M_2}) g$ $\;\;{g}/{{\sqrt{2}}}= (\;\frac{M_2-M_1}{M_1+M_2}) g$ $\; (M_1+M_2) ={\sqrt{2}} M_2-{\sqrt{2}} M_1$ $\;\;\frac{M_1}{M_2}= (\;{{\sqrt{2}}-1}/{{\sqrt{2}}+1})$

$\; {\text{F}}=\;{\Delta {\text{P}}}/{\Delta {\text{t}}}=\;\frac{ {\ mv}-0}{0.1}$ $\;=\;\frac{150 \times 10^{-3} \times 20}{0.1}=30 {\text{N}}$

FBD of ${\text{M}}_1$ :$\; {\text{T}}_1-200=(4+6+10) \times 2$ $\; \therefore {\text{T}}_1=240$