NLM

Ans. (4)
Mass of 10 wagons = $1000 \times 10 = 10000\mathrm{kg}$ So $acc. = \frac{F}{m} = \frac{10^4}{10000} = 1m / s^2$ So on last 4 wagons;
T = (mass × acc.)
= (4 × 1000)(1)
T = 4000 N

Ans. (9)

$V_A = \frac{V_1 + V_2}{2}$[Pulley 1]
$V_1 + V_2 = 2V_A$...(i)
$\frac{V_A + V_1}{2} = V_3$[Pulley 2]
$V_A + V_1 = 0$...(ii)
$\Rightarrow V_1 = -0.6$
$-0.6 + V_2 = 2V_A$
$V_2 = 2 \times 0.6 + 0.6$
$V_2 = 1.8\ m/s$
$V_B = V_2 = 1.8\ m/s$

Ans. (24)
$30 - T = 3aa = 2m/s^2$
$T - 20 = 2aT = 24\ N$

Ans. (1)
$f = \mu N = \mu m a = \frac {1}{2} m \times a = \frac {m a}{2}$
$m g - \frac {m a}{2} = m a _ {1}$
$1 0 - 2 = a _ {1}$
$a _ {1} = 8$
$s = \frac {1}{2} a _ {1} t ^ {2}$

$1 = \frac {1}{2} 8 t ^ {2}$
$t = \frac {1}{2} \sec$

Ans. (67) $f = mg\sin 30^{\circ}$ $f = \mu (mg\cos 30^{\circ} + F) = mg\sin 30^{\circ}$ $\frac{1}{2}\left(20\times \frac{\sqrt{3}}{2} +F\right) = 10$ $5\sqrt{3} +\frac{F}{2} = 10$ $F = 2.68$

Ans. (20) $f_{\mathrm{max}} = \mu N$ $= \frac{1}{4}\times 50\sqrt{3} = 12.5\sqrt{3} = 21.25N$ $50 = 30 + f\Rightarrow f = 20N < f_{\mathrm{max}}$ Hence $f = 20N$

Ans. (4)
Ans. (4) $f = \mu N = 0.5 \times 10 = 5N$ Retardation of block $a_1 = -5 \, m/s^2$ Acceleration of plank $a_2 = 5/2 \, m/s^2$ After t sec velocity of block $10 - 5t$ velocity of plank $\frac{5}{2}t$ $10 - 5t = \frac{5t}{2}$ $10 = \frac{15t}{2}$ ; $t = \frac{4}{3} \sec$

Ans. (5)

For wedge $N \sin 37 = 1 \times 3$
$N \times \frac {3}{5} = 3$
N = 5 N

Ans. (45)
$2 T \cos \theta = \sqrt {2} m g$
$2mg\cos \theta = \sqrt{2} mgs$
$\cos \theta = \frac {\sqrt {2}}{2}$
$\cos \theta = \frac {1}{\sqrt {2}} = \theta = 4 5 ^ {\circ}$

Ans. (4.9)
$7 5 0 - m g \sin 1 5 = 1 0 0 a$
$7 5 0 - 1 0 0 0 \times \frac {2 6}{1 0 0}$
7 5 0 - 2 6 0 = 1 0 0 a

4 9 0 = 1 0 0 a
a = 4. 9

Ans. (3)

The arrangement is shown in the figure.
The separation between the two masses is 2x. Each mass will move in the horizontal direction as shown in the figure.


Let the tension in the string be T. The forces acting at point P and on one of the masses are shown in the figure.
Net force a at point P must equal zero.
$2 T \sin \theta = F \qquad \dots (\mathrm{i})$
Also, For the mass m.
$\begin{array}{l l} T \sin \theta - m g = 0 & \dots (\mathrm{ii}) \\ T \cos \theta = m a & \dots (\mathrm{iii}) \end{array}$
Equation (i) and (iii) give
$a = \frac {F \cot \theta}{2 m} = \frac {F}{2 m} \left(\frac {x}{\sqrt {a ^ {2} - x ^ {2}}}\right)$

Ans. (2)
For motion of a body, down the inclined plane, we have
Force on body = mg sin $45^{\circ} - \mu R$
$m \times \text{acceleration} = mg (\sin 45^{\circ} - \mu mg \cos 45^{\circ})$
$m a = m g \left(\sin 4 5 ^ {\circ} - \mu \cos 4 5 ^ {\circ}\right)$
$m a = \frac {m g}{\sqrt {2}} (1 - \mu)$
$a = \frac {g (1 - \mu)}{\sqrt {2}}\tag{... (i}$

For body A,
$\begin{array}{l} a _ {A} = \frac {g (1 - \mu_ {A})}{\sqrt {2}} \\ a _ {A} = \frac {1 0 (1 - 0 . 2)}{\sqrt {2}} = \frac {8}{\sqrt {2}} = 4 \sqrt {2} m / s ^ {2} \end{array}\tag{... (ii}$
For body B,
$\begin{array}{l} a _ {B} = \frac {g (1 - \mu_ {B})}{\sqrt {2}} \\ a _ {B} = \frac {1 0 (1 - 0 . 3)}{\sqrt {2}} = \frac {7}{\sqrt {2}} = 3. 5 \sqrt {2} m / s ^ {2} \end{array}\tag{... (iii}$
A and B are released simultaneously. They travel for time t. Let A slide by a distance $S_{A}$ . along plane and let B slid by a distance $S_{B}$ along plane till they come on the same line on the inclined plane.
$\therefore S _ {A} = S _ {B} + \sqrt {2}$
$\frac{1}{2} a_{A}t^{2} = \frac{1}{2} a_{B}t^{2} + \sqrt{2}$ , Where $t$ is time taken
$\frac {1}{2} \times (4 \sqrt {2}) \times t ^ {2} = \frac {1}{2} (3. 5 \sqrt {2}) \times t ^ {2} + \sqrt {2},$
From (2) and (3)
$2 t ^ {2} = \frac {3 . 5}{2} t ^ {2} + 1 \Rightarrow 0. 5 t ^ {2} = 2 \Rightarrow t = 2 \sec \quad \dots (\mathrm{iv})$
$\therefore \quad S _ {A} = \frac {1}{2} a _ {A} t ^ {2} = \frac {1}{2} \times (4 \sqrt {2}) (2) ^ {2} = 8 \sqrt {2} m\tag{...(V}$
Hence $S_{A}=8\sqrt{2}m, S_{B}=7\sqrt{2}m, t=2$ second