Motion of Body (or Connected Bodies) on an inclined plane Questions in English

(C) Since the lift is moving with uniform velocity,its acceleration is zero. Therefore,the effective acceleration due to gravity on the block remains '$g$'.
For a block sliding down a frictionless incline,the acceleration is $a = g \sin \theta$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $s = l$:
$l = \frac{1}{2} (g \sin 30^{\circ}) (2)^2 = \frac{1}{2} g (0.5) (4) = g$.
Now,for the incline of $45^{\circ}$,let the time taken be '$t$':
$l = \frac{1}{2} (g \sin 45^{\circ}) t^2$.
Since '$l$' is constant,we equate the two expressions:
$g = \frac{1}{2} g \left(\frac{1}{\sqrt{2}}\right) t^2$.
$1 = \frac{1}{2\sqrt{2}} t^2 \Rightarrow t^2 = 2\sqrt{2} \approx 2.828$.
$t = \sqrt{2.828} \approx 1.68\,s$.

(B) For the system to be at rest (equilibrium),the tension $T$ in the string must balance the weight of $m_{2}$ and the component of weight of $m_{1}$ along the incline.
$T = m_{2}g$
$T = m_{1}g \sin \theta$
Equating the two,we get $m_{2}g = m_{1}g \sin \theta$,which implies $\sin \theta = \frac{m_{2}}{m_{1}} = \frac{3}{5}$.
Since $\sin \theta = \frac{3}{5}$,we have $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$.
The force exerted by the inclined plane on the body of mass $m_{1}$ is the normal force $N$,which balances the component of weight perpendicular to the incline:
$N = m_{1}g \cos \theta$
Substituting the values: $N = 5 \times 10 \times \frac{4}{5} = 40\,N$.

(A) The weight of the block,$W = Mg$,acts vertically downwards.
We can resolve this weight into two rectangular components:
$1$. $A$ component perpendicular to the inclined plane: $Mg \cos \theta$.
$2$. $A$ component parallel to the inclined plane: $Mg \sin \theta$.
Since the block is at rest on the inclined plane,the net force perpendicular to the plane must be zero.
The plane exerts a normal reaction force $N$ on the block,which balances the perpendicular component of the weight.
Therefore,$N = Mg \cos \theta$.
The total force exerted by the plane on the block is the normal reaction force $N$ (assuming a frictionless surface or considering only the normal component as implied by the options).
Thus,the magnitude of the force exerted by the plane on the block is $Mg \cos \theta$.

(D) The force exerted by the plane on the block is the normal reaction force $N$.
When a block of mass $m$ is placed on an inclined plane with an angle of inclination $\theta$,the gravitational force $mg$ acts vertically downwards.
We can resolve the gravitational force into two rectangular components:
$1$. $A$ component perpendicular to the inclined plane: $mg \cos \theta$.
$2$. $A$ component parallel to the inclined plane: $mg \sin \theta$.
Since the block does not move in the direction perpendicular to the plane,the normal force $N$ exerted by the plane must balance the perpendicular component of the gravitational force.
Therefore,$N = mg \cos \theta$.
Thus,the magnitude of the force exerted by the plane on the block is $mg \cos \theta$.

(A) The force $F$ is applied to the pulley. The tension in the string connected to the block $M$ is $F$. The string passes over the pulley,and one end is fixed to the incline. The force $F$ acts at an angle $\theta$ with the incline. The component of $F$ along the incline is $F \cos \theta$. The total force pulling the block up the incline is $T_{total} = F + F \cos \theta$.
For the block to move upward,this force must balance the component of gravity acting down the incline:
$F + F \cos \theta = M g \sin \theta$
$F(1 + \cos \theta) = M g \sin \theta$
$F = \frac{M g \sin \theta}{1 + \cos \theta}$
Using trigonometric identities $\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$ and $1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}$:
$F = \frac{M g (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})}{2 \cos^2 \frac{\theta}{2}}$
$F = M g \tan \frac{\theta}{2}$

(B) Let the mass $m_1 = 4 \, kg$ be on the $60^{\circ}$ incline and $m_2 = 1 \, kg$ be on the $30^{\circ}$ incline.
For the $4 \, kg$ block,the equation of motion is: $m_1 g \sin 60^{\circ} - T = m_1 a \implies 4 \times 10 \times \frac{\sqrt{3}}{2} - T = 4a \implies 20\sqrt{3} - T = 4a \dots (1)$
For the $1 \, kg$ block,the equation of motion is: $T - m_2 g \sin 30^{\circ} = m_2 a \implies T - 1 \times 10 \times \frac{1}{2} = 1a \implies T - 5 = a \dots (2)$
From equation $(2)$,$a = T - 5$. Substituting this into equation $(1)$:
$20\sqrt{3} - T = 4(T - 5)$
$20\sqrt{3} - T = 4T - 20$
$5T = 20\sqrt{3} + 20$
$T = 4(\sqrt{3} + 1) \, N$.

(A) Let the body fall from height $H$. It hits the inclined plane at a height $h$ from the ground,meaning it has fallen a distance of $(H-h)$.
The time taken to fall the distance $(H-h)$ is $t_1 = \sqrt{\frac{2(H-h)}{g}}$.
At this point,the impact is perfectly elastic and the velocity becomes horizontal. The body then falls from height $h$ with an initial vertical velocity of $0$. The time taken to fall this remaining distance $h$ is $t_2 = \sqrt{\frac{2h}{g}}$.
The total time of flight $T$ is $T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}}$.
To find the value of $h$ for which $T$ is maximum,we differentiate $T$ with respect to $h$ and set it to $0$:
$\frac{dT}{dh} = \sqrt{\frac{2}{g}} \left( \frac{1}{2\sqrt{H-h}} \cdot (-1) + \frac{1}{2\sqrt{h}} \right) = 0$.
This implies $\frac{1}{\sqrt{h}} = \frac{1}{\sqrt{H-h}}$,which gives $h = H - h$,or $2h = H$.
Therefore,$\frac{H}{h} = 2$.

(A) Let the radius of the football be $R$ and the radius of the hole be $r$. The diameter of the hole is $2r$. When the plank is tilted at an angle $\theta$,the football rests on the edge of the hole.
At the point of impending motion (rolling),the normal force from the other side of the hole becomes zero.
Let the center of the football be $O$. The football is in contact with the edge of the hole at a point $P$. The distance from the center $O$ to the point of contact $P$ is $R$.
The vertical line passing through the center $O$ makes an angle $\theta$ with the line perpendicular to the plank.
In the triangle formed by the center of the football,the center of the hole,and the point of contact,the distance from the center of the hole to the point of contact is $r$.
Thus,$\sin \theta = \frac{r}{R}$.
Therefore,the maximum angle $\theta$ for which the football does not roll is given by $\sin \theta = \frac{r}{R}$.

(B) To keep the block at rest on the smooth inclined plane,the forces acting on the block must be balanced.
Along the inclined plane,the component of the gravitational force acting downwards is $mg \sin 37^{\circ}$.
The horizontal force $F$ is applied,and its component along the inclined plane acting upwards is $F \cos 37^{\circ}$.
For the block to be in equilibrium,these two forces must be equal:
$F \cos 37^{\circ} = mg \sin 37^{\circ}$
Using the values $\cos 37^{\circ} = \frac{4}{5}$ and $\sin 37^{\circ} = \frac{3}{5}$:
$F \left( \frac{4}{5} \right) = mg \left( \frac{3}{5} \right)$
$F = \frac{3 mg}{4}$

(A) The total mass of the system is $M = 2m + 4m + 6m = 12m$.
The driving force along the incline for the blocks on the right side is $F_1 = (4m + 6m)g \sin 53^{\circ} = 10mg \sin 53^{\circ}$.
The opposing force along the incline for the block on the left side is $F_2 = 2mg \sin 37^{\circ}$.
Using Newton's second law for the system,$F_{net} = Ma$,where $F_{net} = F_1 - F_2$:
$10mg \sin 53^{\circ} - 2mg \sin 37^{\circ} = 12ma$
Substituting the given values $\sin 53^{\circ} = \frac{4}{5}$ and $\sin 37^{\circ} = \frac{3}{5}$:
$10mg \left(\frac{4}{5}\right) - 2mg \left(\frac{3}{5}\right) = 12ma$
$8mg - 1.2mg = 12ma$
$6.8mg = 12ma$
$a = \frac{6.8}{12} g = \frac{68}{120} g = \frac{17}{30} g$.

(A) For the given system,the equation of motion is derived from the net force divided by the total mass.
$a = \frac{M_2 g \sin \theta}{M_1 + M_2}$
When the mass of $M_1$ is doubled $(M_1' = 2M_1)$ and the mass of $M_2$ is halved $(M_2' = M_2/2)$,the new acceleration $a'$ is:
$a' = \frac{M_2' g \sin \theta}{M_1' + M_2'} = \frac{(M_2/2) g \sin \theta}{2M_1 + M_2/2}$
Multiplying the numerator and denominator by $2$:
$a' = \frac{M_2 g \sin \theta}{4M_1 + M_2}$
Now,express $a'$ in terms of $a$:
$a' = \left( \frac{M_2 g \sin \theta}{M_1 + M_2} \right) \times \left( \frac{M_1 + M_2}{4M_1 + M_2} \right) = a \left( \frac{M_1 + M_2}{4M_1 + M_2} \right)$

(B) For an object moving at a constant speed on a frictionless inclined plane, the net force acting on it must be zero.
When the object is pulled up the plane, the applied force $F$ must balance the component of the gravitational force acting down the plane.
The component of the weight $W = mg$ acting down the inclined plane is $mg \sin \theta$.
Given:
Applied force $F = 500 \,N$
Angle of inclination $\theta = 30^{\circ}$
Since the object moves at a constant speed, the forces are in equilibrium:
$F = mg \sin 30^{\circ}$
$500 \,N = mg \cdot \frac{1}{2}$
$mg = 500 \,N \times 2$
$mg = 1000 \,N$
Therefore, the weight of the object is $1000 \,N$.

(B) Let $m_2 = 300 \ g$ and $m_1 = 200 \ g$. The angle of inclination is $\theta = 30^{\circ}$.
For the block of mass $m_2$,the equation of motion is: $m_2 g \sin \theta - T = m_2 a$ $(i)$
For the block of mass $m_1$,the equation of motion is: $T - m_1 g \sin \theta = m_1 a$ (ii)
Adding equations $(i)$ and (ii),we get:
$(m_2 - m_1) g \sin \theta = (m_1 + m_2) a$
$a = \frac{(m_2 - m_1) g \sin \theta}{m_1 + m_2}$
Substituting the values: $a = \frac{(300 - 200) g \sin 30^{\circ}}{300 + 200}$
$a = \frac{100 \times g \times 0.5}{500} = \frac{50}{500} g = \frac{1}{10} g$
$a = 0.1 g = 10 \% \text{ of } g$.

(C) The length of an inclined plane of height $h$ and angle $\theta$ is $L = \frac{h}{\sin \theta}$.
The acceleration of a block sliding down a smooth inclined plane is $a = g \sin \theta$.
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $s = L$:
$L = \frac{1}{2} (g \sin \theta) t^2 \implies \frac{h}{\sin \theta} = \frac{1}{2} g \sin \theta \ t^2$.
Thus,$t = \sqrt{\frac{2h}{g \sin^2 \theta}} = \frac{1}{\sin \theta} \sqrt{\frac{2h}{g}}$.
Given $h = 20 \ m$ and $g = 10 \ m \ s^{-2}$,we have $\sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 20}{10}} = 2 \ s$.
For plane $A$ $(\theta_1 = 30^{\circ})$: $t_1 = \frac{1}{\sin 30^{\circ}} \times 2 = \frac{1}{0.5} \times 2 = 4 \ s$.
For plane $B$ $(\theta_2 = 60^{\circ})$: $t_2 = \frac{1}{\sin 60^{\circ}} \times 2 = \frac{1}{\sqrt{3}/2} \times 2 = \frac{4}{\sqrt{3}} \ s$.
Therefore,$t_1 - t_2 = 4 - \frac{4}{\sqrt{3}} = 4 \left( 1 - \frac{1}{\sqrt{3}} \right) = 4 \left( \frac{\sqrt{3}-1}{\sqrt{3}} \right) \ s$.

(B) The free body diagrams of the blocks are shown in the figure.
At equilibrium,the net force on a body is zero.
For block $A$ hanging vertically:
$m_A g = k x$
$\Rightarrow k = \frac{m_A g}{x}$
For block $B$ on the inclined plane:
$m_B g \sin 30^{\circ} = k x^{\prime}$
$\Rightarrow x^{\prime} = \frac{m_B g \sin 30^{\circ}}{k} = \frac{m_B g \sin 30^{\circ}}{(m_A g / x)} = \frac{m_B}{m_A} x \sin 30^{\circ}$
Since both blocks are made of the same material,mass is proportional to volume $(m = \rho V)$:
$\Rightarrow x^{\prime} = \frac{V_B}{V_A} x \sin 30^{\circ} = \frac{0.75}{0.25} \times 0.2 \times \sin 30^{\circ}$
$\Rightarrow x^{\prime} = 3 \times 0.2 \times 0.5 = 0.3 \ m$
The total distance of the block from the top is the sum of the unstretched length $l$ and the extension $x^{\prime}$:
$d = l + x^{\prime} = 1.0 \ m + 0.3 \ m = 1.3 \ m$.

(A) Let the mass of each block be $m$. The forces acting along the incline are $mg \sin 60^{\circ}$ and $mg \sin 30^{\circ}$.
Since $mg \sin 60^{\circ} > mg \sin 30^{\circ}$,the system accelerates such that the block on the $60^{\circ}$ incline moves down.
For the block on the $60^{\circ}$ incline: $mg \sin 60^{\circ} - T = ma$ --- $(i)$
For the block on the $30^{\circ}$ incline: $T - mg \sin 30^{\circ} = ma$ --- (ii)
Adding $(i)$ and (ii): $mg(\sin 60^{\circ} - \sin 30^{\circ}) = 2ma$
$a = \frac{g}{2} \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \frac{(\sqrt{3}-1)g}{4}$.
The acceleration vectors for the blocks are $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.
The acceleration of the centre of mass is $\vec{a}_{cm} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.
$\vec{a}_{cm} = \frac{a}{2} [(\cos 60^{\circ} - \cos 30^{\circ}) \hat{i} - (\sin 60^{\circ} + \sin 30^{\circ}) \hat{j}]$.
Using $\cos 60^{\circ} = 1/2, \cos 30^{\circ} = \sqrt{3}/2, \sin 60^{\circ} = \sqrt{3}/2, \sin 30^{\circ} = 1/2$:
$|\vec{a}_{cm}| = \frac{a}{2} \sqrt{(\frac{1-\sqrt{3}}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \frac{a}{2} \sqrt{\frac{1+3-2\sqrt{3} + 3+1+2\sqrt{3}}{4}} = \frac{a}{2} \sqrt{\frac{8}{4}} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Substituting $a$: $|\vec{a}_{cm}| = \frac{(\sqrt{3}-1)g}{4\sqrt{2}}$.

(D) The total mass of the system is $M = (8 + 5 + 3) \,kg = 16 \,kg$.
Since the blocks are connected and move together,the entire system can be treated as a single body of mass $M = 16 \,kg$ moving up the incline with acceleration $a$.
The external force applied is $F = 104 \,N$.
The component of the total weight acting down the incline is $Mg \sin 30^{\circ}$.
Applying Newton's second law to the system:
$F - Mg \sin 30^{\circ} = Ma$
$104 - 16 \times 10 \times \sin 30^{\circ} = 16a$
$104 - 160 \times 0.5 = 16a$
$104 - 80 = 16a$
$24 = 16a$
$a = \frac{24}{16} = 1.5 \,m / s^2$.
Thus,the acceleration of the blocks is $1.5 \,m / s^2$.