# Newton's Laws of Motion

## Concept Of Physics

### H C Verma

1   A block of mass $2 kg$ placed on a long frictionless horizontal table is pulled horizontally by a constant force $F$. It is found to move $10 m$ in the first two seconds. Find the magnitude of $F$.

##### Solution :

2   A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it ?

##### Solution :

$u$ = $\frac{40 km}{hr}$ = $\frac{40000}{3600}$ = $11.11$ $\frac{m}{s}$ $\\$ $m$ = $2000$ $kg$ : $v$ = $0$ : $s$ = $4m$ $\\$ acceleration $'a'$ = $\frac{v^2-u^2}{2s}$ = $\frac{0^2-(11.11)^2}{2 x 4}$ = -$\frac{123.43}{8}$ = $-15.42$ $\frac{m}{s^2}$ (deceleration) $\\$ So, braking force = $F$ = $ma$ = $2000$ x $15.42$= $30840$ = $3.08$ $10^4$ $N$ (Ans)

3   A particle of mass 50 g moves on a straight line. The variation of speed with time is shown in figure (5-E1). Find the force acting on the particle at t = 2, 4 and 6 seconds.

##### Solution :

$m$= $50 g$ = $5$ x $10^{-2}$ $kg$ $\\$ Slope of $OA$ = $Tan$$\theta \frac{AD}{OD} = \frac{15}{3} = \frac{5m}{s^2} \\ So, at t = 2 sec acceleration is \frac{5m}{s^2} \\ Force = ma = 5 x 10^{-2} x 5 = 0.25 N along the motion \\ At t = 4 sec \\ slope of AB = 0, acceleration = 0 [tan$${\theta{ ^o}}$ = $0$] $\\$ Force = $0$ $\\$ At $t$ = $6$ $sec$, acceleration = slope of $BC$ $\\$ In $\Delta$$BEC = tan$$\theta$ = $\frac{BE}{EC}$ = $\frac{15}{3}$ = $5$ $\\$ Slope of $BC$ = $tan$ ($180^o$ - $\theta$) = -$tan\theta$ = -$5$ $\frac{m}{s^2}$ (deceleration) $\\$ Force = $ma$ = $5$ x $10^{-2}$ $5$ = $0.25$ $N$. Opposite to the motion.

4   In a TV picture tube electrons are ejected from the cathode with negligible speed and reach a velocity of $5$ x $10 6 m/s$ in travelling one centimeter. Assuming straight line motion, find the constant force exerted on the electron. The mass of the electron is $9.1$ x $10$ - $31 kg$

##### Solution :

Initial Velocity $u$ = $0$ (negligible)$\\$ $v$ = $5$ x $10^6$ $\frac{m}{s}$ $\\$ $s$ = $1$ $cm$ = $1$ x $10^{-2}$ $m$ $\\$ acceleration $a$ = $\frac{v^2 - u^2}{2s}$= $\frac{(5 \times 6^2) - 0}{2 \times 1 \times 10^{-2}}$= $\frac{25 \times 10^2}{2 \times 10^{-2}}$= $12.5$ x $10^{14}$ m$s^{-2}$ $\\$ $F$ = $ma$ = $9.1$ x $10^{-31}$ x $12.5$ x $10^{14}$ = $113.75$ x $10^{-17}$ = $1.1$ x $10^{-15}$ $N$

5   Two blocks $A$ and $B$ of mass $m^A$, and $m^B$ respectively are kept in contact on a frictionless table. The experimenter pushes the block $A$ from behind so that the blocks accelerate. If the block $A$ exerts a force $F$ on the block $B$, what is the force exerted by the experimenter on $A$ ?

##### Solution :

Let $F$ $\rightarrow$ contact force between $m_A$ & $m_B$. $\\$ And, $f$ $\rightarrow$ force exerted by experimenter. $\\$ $F$ + $m_A{a}$ - $f$ = $0$ , $m{_B}a$ - $f$ = $0$ $\\$ $\Rightarrow$ $F$ = $f$ - $m_Aa$ ......$(i)$ $\\$ $\Rightarrow$ $F$ = $m_Ba$......$(ii)$ $\\$ From eqn $(i)$ and eqn $(ii)$ $\\$ $\Rightarrow$ $f$ - $m_Aa$ = $m_Ba$ $\Rightarrow$ $f$ = $m_Ba$ + $m_aA$ $\Rightarrow$ $f$ = $a$ $(m_A+m_B)$ $\\$ $f$ = $\frac{F}{m_B}$ ($m_B$ + $m_A$) = $F$ $\big(1+ \frac{m_A}{m_B} \big)$ [because $a$ = $\frac{F}{m_B}$] $\\$ $\therefore$ The force exerted by the experimenter is $F$ $\big(1+\frac{m_A}{m_B}\big)$

6   A block of mass $0.2 kg$ is suspended from the ceiling by a light string. A second block of mass $0.3 kg$ is suspended from the first block through another string. Find the tensions in the two strings. Take $g$ = $10$ $\frac{m}{s^{2}}$

##### Solution :

$g$ = $10$ $\frac{m}{s^{2}}$ $\\$ $T$ - $0.3g$ = $0$ $\Rightarrow$ $T$ = $0.3$ $g$ = $0.3$ x $10$ = $3$ $N$ $\\$ $T_1$ = ($0.2g$ + $T$) = $0$ $\Rightarrow$ $T_1$ = $0.2g$ + $T$ = $0.2$ x $10$ + $3$ = $5$ $N$

7   Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force $F$. Find the tension in the string joining the blocks.

##### Solution :

$T$ + $ma$ - $F$ = $0$ $\\$ ($T$ - $ma$ = $0$ $\Rightarrow$ $T$ = $ma$ - ($Eqn.1$) $\\$ $\Rightarrow$ $F$ = $T$ + $ma$ $\Rightarrow$ $F$ = $T$ + $T$ ....($from Eqn.1$)$\\$ $\Rightarrow$ $2T$ = $F$ $\Rightarrow$ $T$ = $\frac{F}{2}$

8   Raindrops of radius $1 mm$ and mass $4 mg$ are falling with a speed of $30$ $\frac{m}{s}$ on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.

##### Solution :

$r$ = $1$ $mm$ = $10^{-3}$ $\\$ $'m'$ = $4$ $mg$ = $4$ x $10^{-6} kg$ $\\$ $8$ = $10^{-3} m$. $\\$ $v$ = $0$ $\\$ $u$ = $30 \frac{m}{s}$ $\\$ So, $a$ = $\frac{v^2-u^2}{2s}$ = $\frac{{-30}\times30}{2 \times {10^{-3}}}$ = $-4.5$ x $10^{5}$ $\frac{m}{s^2}$ (decelerating) $\\$ Taking magnitude only deceleration is $4.5$ x $10^{6}$ $\frac{m}{s^2}$ $\\$ So, force $F$ = $4$ x $10^{-6}$ x $4.5$ X $10^{6}$ = $1.8$ $N$ $\\$

9   A particle of mass $0.3$ $kg$ is subjected to a force $F$ = $-kx$ with $k$ = $15$ $\frac{N}{m}$. What will be its initial acceleration if it is released from a point = $20$ $cm$ ?

##### Solution :

$x$ = $20$ $cm$ = $0.2$ $m$, $k$ = $\frac{15N}{m}$, $m$ = $0.3$ $kg$ $\\$ Acceleration $a$ = $\frac{F}{m}$ = $\frac{-kx}{x}$ = $\frac{-15(0.2)}{0.3}$ = $\frac{3}{0.3}$ = $-10\frac{m}{s^2}$ {deceleration) $\\$ So, the acceleration is $10\frac{m}{s^2}$ opposite to the direction of motion.

10   Both the springs shown in figure (5-E2) are un stretched. If the block is displaced by a distance $x$ and released, what will be the initial acceleration?

##### Solution :

Let the block $m$ towards left through displacement $x$ $\\$ $F_1$ = $k_1$ $x$ (compressed) $\\$ $F_2$ = $k_2$ $x$ (expanded) $\\$ They are in same direction. $\\$ Resultant $F$ = $F_1$ + $F_2$ $\Rightarrow$ $F$ = $k_1x$ + $k_2x$ $\Rightarrow$ $F$ = $x$($k_1 + k_2$) $\\$ So, $a$ = acceleration = $\frac{F}{m}$ = $\frac{x(k_1+K_2)}{m}$ opposite to the displacement.

11   A man has fallen into a ditch of width $d$ and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth $h$.

##### Solution :

a) at any depth let the ropes make angle $\theta$ with the vertical From the free body diagram $\\$ $F$ $cos\theta$+ $F$ $cos\theta$ - $mg$ = $0$ $\\$ $\Rightarrow$ $2F$ $cos\theta$ = $mg$ $\Rightarrow$ $F$ = $\frac{mg}{2cos\theta}$ $\\$ As the man moves up, $\theta$ increases i.e, $cos$ $\theta$ decreases. Thus $F$ increases. $\\$

b) When the man is at depth $h$ $\\$ $cos$ $\theta$ = $\frac{h}{\sqrt{\frac{d}{2}{^2}+h^2}}$ $\\$ Force = $\frac{mg}{h}$ = $\frac{mg}{4h}$$\sqrt{d^2+4^2} \\ \sqrt{{\frac{d^2}{4}}+h^2} 12 The elevator shown in figure (5-E5) is descending with an acceleration of \frac{2m}{s^2} . The mass of the block A is 0.5 kg. What force is exerted by the block A on the block B ? ##### Solution : From the free diagram \\ \therefore R + 0.5 x 2 - w = 0 \\ \Rightarrow = w - 0.5 x 2 \\ = 0.5 (10-2) = 4N \\ So, the force exerted by the block A on the block B is 4N 13 A pendulum bob of mass 50 g is suspended from the ceiling of an elevator. Find the tension in the string if the elevator (a) goes up with acceleration \frac{1.2m}{s^2}, (b) goes up with deceleration \frac{1.2 m}{s^2}, (c) goes up with uniform velocity, (d) goes down with acceleration \frac{1.2 m}{s^2}, (e) goes down with deceleration \frac{1.2 m}{s^2} and (f) goes down with uniform velocity. ##### Solution : a)The tension in the string is found out for the different conditions from the body diagram shown below. \\ T - (W + 0.06 \times 1.2) = 0 \\ \Rightarrow = 0.05 x 9.8 + 0.05 x 1.2 \\ = 0.56 N. \\ b) \therefore T + 0.05 x 1.2 - 0.05 x 9.8 = 0 \\ \Rightarrow T = 0.05 x 9.8 - 0.5 x 1.2 \\ = 0.43 N. \\ c) When the elevator makes uniform motion \\ T - W = 0 \\ \Rightarrow T = W - 0.05 x 9.8 \\ = 0.49 N \\ d) T + 0.05 x 1.2 - W = 0 \\ \Rightarrow T = W - 0.05 x 1.2 \\ = 0.43 N \\ e) T - (W+0.05\times) = 0 \Rightarrow T = W + 0.05 x 1.2 \\ = 0.55 N \\ f) When the elevator goes down with the uniform velocity acceleration = 0 \\ T-W=0 \\ \Rightarrow$$T$=$W$ = $0.05$ x $9.8$ $\\$ =$0.49$ $N$

14   A small block $B$ is placed on another block $A$ of mass $5 kg$ and length $20 cm$. Initially the block $B$ is near the right end of block $A$ (figure 5-E3). $A$ constant horizontal force of $10 N$ is applied to the block $A$. All the surfaces are assumed friction less. Find the time elapsed before the block $B$ separates from $A$.

##### Solution :

$m$ = $5 kg$ of mass $A$ $\\$ $ma$ = $10$ $N$ $\\$ $\Rightarrow$ $a$ $\frac{10}{5}$ = $2\frac{m}{s^2}$ $\\$

As there is no friction between $A$ and $B$, when the block $A$ moves, Block $B$ remains at rest in its motion $\\$ initial velocity of $A$ = $u$ = $0$ $\\$ Distance to cover so that B seperate out $s$ = $0.2m$ $\\$ Acceleration $a$ = $\frac{2m}{s^2}$ $\\$ $\therefore$ $s$ = $ut$ + $\frac{1}{2}$ $at^2$ $\\$ $\Rightarrow$ $0.2$ = $0$ + ($\frac{1}{2})$ x $2$ x $t^2$ $\Rightarrow$ $t^2$ = $0.2$ $\Rightarrow$ $t$ = $0.44$ sec $\Rightarrow$ $t$ = $0.45$ sec

15   A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are $72 kg$ and $60 kg$. Assuming that the magnitudes of the acceleration and the deceleration are the same, find $(a)$ the true weight of the person and $(b)$ the magnitude of the acceleration. Take $g$ = $\frac{9.9m}{s^2}$.

##### Solution :

When the elevator is accelerating upwards, maximum weight will be recorded. $\\$ $R$ - ($W$ + $ma$) = $0$ $\\$ $\Rightarrow$ $R$ = $W$ +$ma$ = $m(g+a)$ max.wt $\\$ When decelerating upwards, maximum weight will be recorded.$\\$ $R$ + $ma$ - $W$ = $0$ $\\$ $\Rightarrow$ $R$ = $W$ - $ma$ = $m(g-a)$ $\\$ So, $m(g+a)$ = $72$ x $9.9$.....(1) $\\$ $m(g-a)$ = $60$ x $9.9$.....(2) $\\$ Now, $mg$ + $ma$ = $72$ x $9.9$ $\Rightarrow$ $mg$ - $ma$ = $60$ x $9.9$ $\\$ $\Rightarrow$ $2mg$ = $1306.8$ $\\$ $\Rightarrow$ $m$ = $\frac{1306.8}{2 \times 9.9}$ = $66$ $Kg$ $\\$ So the true weight of the man is $66$ $kg$ $\\$ Again, to find the acceleration, $mg$ + $ma$ = $72$ x $9.9$ $\\$ $\Rightarrow$ $a$ = $\frac{72 \times 9.9 - 66 \times 99}{66}$ = $\frac{9.9}{11}$ = $0.9$ $\frac{m}{s^2}$.

16   Find the reading of the spring balance shown in figure (5-E6). The elevator is going up with an acceleration of $\frac{g}{10}$, the pulley and the string are light and the pulley is smooth.

##### Solution :

Let the acceleration of the 3 kg mass relative to the elevator is $'a'$ in the downward direction. $\\$ As, shown in the free body diagram $\\$ $T$ = $1.5$ $g$ = $1.5(\frac{g}{10})$ = $1.5$ $a$ = $0$.....from figure $(1)$ $\\$ and, $T$ - $3g$ - $3(\frac{g}{10})$ + $3a$ = $0$.....from figure $(2)$ $\\$ $\Rightarrow$ $T$ = $1.5 g$ + $1.5(\frac{g}{10})$ + $1.5a$ .....$(i)$ $\\$ And $T$ = $3g$ + $3(\frac{g}{10})$ - $3a$.....$(ii)$ $\\$ Equation $(i)$ x $2$ $\Rightarrow$ $3g$ +$3(\frac{g}{10})$ + $3a$ = $2T$ $\\$ Equation $(ii)$ x $1$ $\Rightarrow$ $3g$ +$3(\frac{g}{10})$ - $3a$ = $T$ $\\$ Subtracting the above two equations we get, $T$ = $6a$ $\\$ Subtracting $T$ = $6a$ in equation $(ii)$ $\\$ $6a$ = $3g$ + $3(\frac{g}{10})$ - $3a$ $\\$ $\Rightarrow$ $9a$ = $\frac{33g}{10}$ $\Rightarrow$ $a$ = $\frac{(9.8)33}{10}$ = $32.34$ $\Rightarrow$ $a$ = $3.59$ $\\$ $\therefore$ $T$ = $6a$ = $6$ x $3.59$ = 21.55 $\\$ $T^1$ = $2T$ = $2$ x $21.55$ = $43.1$ $N$ cut is $T_1$ shown in spring. $\\$ Mass = $\frac{wt}{g}$ = $\frac{43.1}{9.8}$ = $4.39$ = $4.4$ kg

17   A block of $2 kg$ is suspended from the ceiling through a massless spring of spring constant $k$ = $\frac{100N}{m}$ What is the elongation of the spring ? If another $1 kg$ is added to the block, what would be the further elongation ?

##### Solution :

Given $m$ = $2kg$, $k$ = $\frac{100N}{m}$ $\\$ From the free body diagram, $kl$ - $2g$ = $0$ $\Rightarrow$ $kl$ = $2g$ $\\$ $\Rightarrow$ $l$ = $\frac{2g}{k}$ = $\frac{2 \times 9.8}{100}$ = $\frac{19.6}{100}$ = $0.196$ = $0.2$ $m$ $\\$ Suppose further elongation when $1$ $kg$ block is added to be $x$, $\\$ Then $k(1 + x)$ = $3g$ $\\$ $\Rightarrow$ $kx$ = $3g$ - $2g$ = $g$ = $9.8$ $N$ $\\$ $\Rightarrow$ $x$ = $\frac{9.8}{100}$ = $0.098$ = $0.1$ $m$

18   Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of $\frac{2.0m}{s^2}$. Find the elongations.

##### Solution :

$a$ = $\frac{2m}{s^2}$ $\\$ $kl$ - $(2g + 2a)$ = $0$ $\\$ $\Rightarrow$ $kl$ = $2g$ + $2a$ $\\$ = $2$ x $9.8$ + $2$ x $2$ = $19.6$ + $4$ $\\$ $\Rightarrow$ $l$ = $\frac{23.6}{100}$ = $0.0236$ = $0.24$ $\\$ $\\$

When $1kg$ body is added total mass $(2+1)kg$ = $3kg$. $\\$ elongation be $l_1$, $\\$ $kl_1$ = $3g$ + $3a$ = $3$ x $9.8$ + $6$ $\\$ $\Rightarrow$ $l1$ = $\frac{33.4}{100}$ = $0.0334$ = $0.36$ $\\$ Further elongation = $l_1$ - $1$ = $0.36$ - $0.12$ $m$

19   The force of buoyancy exerted by the atmosphere on a balloon is $B$ in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of velocity and is proportional to it. The balloon carries a mass $M$ and is found to fall down near the earth's surface with a constant velocity $V$. How much mass should be removed from the balloon so that it may rise with a constant velocity $V$ ?

For the balloon to rise with a constant velocity $v$, (upward) let the mass be $m$ $\\$ Here, $B$ - ($mg + kv)$ = $0$ .....$(ii)$ $\\$ $\Rightarrow$ $B$ = $mg$ + $kv$ $\\$ $\Rightarrow$ $m$ = $\frac{B-kw}{g}$ $\\$ So, amount of mass that should be removed = $M$ - $m$ $\\$ =$\frac{B+kv}{g}$ - $\frac{B-kv}{g}$ = $\frac{B+kv-B+kv}{g}$ = $\frac{2kv}{g}$ = $\frac{2(Mg-B)}{G}$ = $2(M-(\frac{B}{g})$

Let, the air resistance force is $F$ and Buoyant force is $B$ $\\$ Given that $\\$ $F_a$ $\infty$ $v$ where $v$ $\rightarrow$ velocity $\\$ $\Rightarrow$ $F_a$ = $kv$, where $k$ $\rightarrow$ proportionality constant. $\\$ When the balloon is moving downward, $\\$ $B$ + $kv$ = $mg$ .....$(i)$ $\\$ $\Rightarrow$ $M$ = $\frac{B+kv}{g}$ $\\$

##### Solution :

Let, the air resistance force is $F$ and Buoyant force is $B$ $\\$ Given that $\\$ $F_a$ $\infty$ $v$ where $v$ $\rightarrow$ velocity $\\$ $\Rightarrow$ $F_a$ = $kv$, where $k$ $\rightarrow$ proportionality constant. $\\$ When the balloon is moving downward, $\\$ $B$ + $kv$ = $mg$ .....$(i)$ $\\$ $\Rightarrow$ $M$ = $\frac{B+kv}{g}$ $\\$

For the balloon to rise with a constant velocity $v$, (upward) let the mass be $m$ $\\$ Here, $B$ - ($mg + kv)$ = $0$ .....$(ii)$ $\\$ $\Rightarrow$ $B$ = $mg$ + $kv$ $\\$ $\Rightarrow$ $m$ = $\frac{B-kw}{g}$ $\\$ So, amount of mass that should be removed = $M$ - $m$ $\\$ =$\frac{B+kv}{g}$ - $\frac{B-kv}{g}$ = $\frac{B+kv-B+kv}{g}$ = $\frac{2kv}{g}$ = $\frac{2(Mg-B)}{G}$ = $2(M-(\frac{B}{g})$

20   An empty plastic box of mass $m$ is found to accelerate up at the rate of $\frac{g}{6}$ when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of $\frac{g}{6}$ ?

When the box is accelerating upward, $\\$ $U$ - $mg$ - m($\frac{g}{6}$ = $0$ $\\$ $\Rightarrow$ $U$= $mg$ + $\frac{mg}{6}$ = $m(g + (\frac{g}{6}))$ = $7$ $\frac{mg}{7}$ .....$(i)$ $\\$ $\Rightarrow$ $m$ = $\frac{6U}{7g}$ $\\$ Whenit is accelerating downward, let the required mass be $M$. $\\$ $U$- $Mg$ + $\frac{Mg}{6}$ = $0$ $\\$ $\Rightarrow$ $U$ = $\frac{6Mg-mg}{6}$ = $\frac{5Mg}{6}$ $\Rightarrow$ $M$ = $\frac{6U}{5g}$ $\\$ Mass to be added = $M$ - $m$ = $\frac{6U}{5g}$-$\frac{6U}{7g}$ = $\frac{6U}{g}$ $\big($$\frac{1}{5}-\frac{1}{7}$$\big)$ $\\$ = $\frac{6U}{g}$ $\big($$\frac{2}{35}$$\big)$ = $\frac{12}{35}$ $\big($$\frac{U}{g}$$\big)$ $\\$ = $\frac{12}{35}$ $\big($$\frac{7mg}{6} \times \frac{1}{g} \big) from (i) \\ = \frac{2}{5} m.\\ \therefore The mass to be added is \frac{2m}{5} 21 A force \overrightarrow{F}= \overrightarrow{v} x \overrightarrow{A} is exerted on a particle in addition to the force of gravity, where \overrightarrow{v} is the velocity of the particle and \overrightarrow{A} is a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues to move undeflected with a constant velocity ? ##### Solution : Given that \overrightarrow{F} = \overrightarrow{u} x \overrightarrow{A} and \overrightarrow{mg} act on the particle. \\ For the particle to move undeflected with constant velocity, net force should be zero. \\ \therefore (\overrightarrow{u} x \overrightarrow{A}) + (\overrightarrow{mg}) = 0 \\ \therefore (\overrightarrow{u} x \overrightarrow{A}) = -\overrightarrow{mg} \\ Because, (\overrightarrow{u} x \overrightarrow{A}) is perpendicular to the plane containing \overrightarrow{u} and \overrightarrow{A}, \overrightarrow{u} should be in the x-z plane \\ Again ,u A sin\theta = mg \\ \therefore u = \frac{mg}{A sin\theta } \\ u will be minimum , when sin \theta = 1 \Rightarrow \theta = 90^o \\ \therefore u_{min} = \frac{mg}{A} along Z-axis 22 In a simple Atwood machine, two unequal masses m1, and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5-E7) m, = 300 g and m2 = 600 g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley ##### Solution : m_1 = 0.3 kg, m_2 = 0.6 kg \\ T - (m_{1}g +m_{1}a) = 0 ....(i) \Rightarrow T= m_{1}g +m_{1}a \\ T + (m_{2}a -m_{2}g) = 0 ....(ii) \Rightarrow T= m_{2}g +m_{2}a \\ From equation (i) and (ii) m_{1}g +m_{1}a + m_{2}a +m_{2}g = 0, from (i) \Rightarrow a(m_1+ m_2) = g(m_2-m_1) \\ \Rightarrow a =f \big($$\frac{m_2-m_1}{m_1+m_2}$$\big) = 9.8$$\big($$\frac{0.6-0.3}{0.6+0.3}$$\big)$ = $3.266$ $ms^2$.$\\$ $a)$ $t$ = $2$ sec acceleration = $3.266$ $ms^{-2}$ $\\$ Initial velocity $u$ = $0$ $\\$ So, distance travelled by the body is, $\\$ $S$ = $ut$ + $\frac{1}{2}at^2$ $\Rightarrow$ $0$ + $\frac{1}{2}$$(3.266) 2^2 = 6.5 m. \\ b) From (i) T = m_1(g+a) = 0.3 (9.8 + 3.26) = 3.9 N \\ c) The force exerted by the clamp on the pully is giiven by F - 2T = 0 \\ F = 2T = 2 x 3.9 = 7.8 N 23 Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0s after the system is set into motion. Find the time elapsed before the string is tight again. ##### Solution : a = 3.26 \frac{m}{s^2} \\ T = 3.9N \\ After 2 sec mass m_1 the velocity V = u + at = 0 + 3.26 x 2 = \frac{6.52m}{s} upward.\\ At this time m_2 is moving \frac{6.52m}{s} downward. \\ At time 2 sec, m_2 stops for a moment, But m_1 is moving upward with velocity \frac{6.52m}{s} \\ Itwill continue to move till final velocity (at highest point) because zero \\ Here , v = 0 ; u = 6.52 \\ A = -g = \frac{-9.8m}{s^2} [moving upward m_1] \\ V = u + at \Rightarrow 0 = 6.52 + (-9.8)t \\ \Rightarrow t = \frac{6.52}{9.8} = 0.66 = \frac{2}{3} sec \\ During the period \frac{2}{3}, m_2 mass also starts moving downward. So the string becomes tight after a time of \frac{2}{3} sec. 24 Figure (5-E8) shows a uniform rod of length 30 cm having a mass of 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light. ##### Solution : Mass per unit length \frac{3}{30} \frac{kg}{cm} = 0.10 \frac{kg}{cm} \\ Mass of 10 cm part = m_1 = 1 kg \\ Mass of 20 cm part = m_2 = 2 kg \\ Let, F = contact force between them, \\ From the free body diagram \\ F - 20 - 10 = 0 .....(i) \\ And, 32 - F - 2a = 0 .....(ii) From eqn (i) and (ii) \\ 3a - 12 = 0 \Rightarrow a = \frac{12}{3} = \frac{4m}{s^2} \\ Contact force F = 20 + 1a = 20 + 1 x 4 = 24 N. 25 Consider the situation shown in figure (5-E9). All the surfaces are frictionless and the string and the pulley are light. Find the magnitude of the acceleration of the two blocks. ##### Solution : Sin\theta_1 = \frac{4}{5} \\ gSin\theta_1-(a+T) = 0 \\ T - g Sin\theta_2-a = 0 \\ Sin\theta_2 = \frac{3}{5} \Rightarrow g$$Sin\theta_1$ = $a$ +$T$ .......$(i)$ $\\$ $\Rightarrow$ $T$ = $g$ $sin\theta_2$ + $a$ .....$(ii)$ $\\$ $\Rightarrow$ $T$ + $a$ - $g$ $sin\theta_2$ = $0$ $\\$ From eqn $(i)$ and $(ii)$ , $g$ $sin\theta_2$ + $a$ + $a$ - $g$ $sin\theta_1$ = $0$ $\\$ $\Rightarrow$ $2a$ = $g$ $sin\theta_1$ - $g$ $sin\theta_2$ = g$\big($$\frac{4}{5}-\frac{3}{5}$$\big)$ = $\frac{g}{5}$ $\\$ $\Rightarrow$ $a$ = $\frac{g}{5}$x$\frac{1}{2}$ = $\frac{g}{10}$

26   A constant force $F=\frac{m_2}{g}$ is applied on the block of $E$ mass $m$, as shown in figure (5-E10). The string and the pulley are light and the surface of the table is smooth. Find the acceleration of $m$.

From the above free body diagram $\\$ $M_1{A}$ + $F$ - $T$ = $0$ $\Rightarrow$ $T$ = $m_1{a}$ + $F$ .....$(i)$ $\\$ From the above free diagram $\\$ $m_2{a}$ + $T$ - $m_2{g}$ = $0$ .....$(ii)$ $\\$ $\Rightarrow$ $m_2{a}$ +$m_1{a}$ + $F$ - $m_2{g}$ = $0$ (from $(i)$ $\\$ $\Rightarrow$$a(m_1+m_2)+m_2\frac{g}{2} - m_2g = 0 {because f = m^2\frac{g}{2}} \\ \Rightarrow a(m_1+m_2) -m_2{g} = 0 \\ \Rightarrow a(m_1+m_2) =\frac{m_2{g}}{2} \\ \Rightarrow a = \frac{m_2}{2(m_1 = m_2{g})} 27 In figure (5-E11) m_1= 5 kg, m_2 = 2 kg and F = 1N . Find the acceleration of either block. Describe the motion of m_1, if the string breaks but F continues to act. ##### Solution : From the above free body diagram \\ T + m_1{a} - m(m_1{g}+F) = 0 \\ From the above free body diagram \\ T- (m_2{g}+F+m_2{a}) = 0 \Rightarrow T = m_1{g}+F-m_1{a}\Rightarrow T=5g+1-5a.....(i) \\ \Rightarrow T = m_2{g}+F+m_2{a}\Rightarrow T=2g+1+2a.....(ii) \\ From the eqn (i) and (ii) \\ 5g + 1 - 5a = 2g + 1 +2a \Rightarrow 3g - 7a=0\Rightarrow 7a = 3g \\ \Rightarrow a=\frac{3g}{7} = \frac{29.4}{7} = \frac{4.2m}{s^2} [g=\frac{9.8m}{s^2}] \\ a) acceleration of block is \frac{4.2m}{s^2} \\ b) After the string breaks m_1 move downward with force F acting downward. \\ m_a = F +m_2{g}=(1+5g)=5(g+0.2) \\ Force = 1N, acceleration = \frac{1}{5} = \frac{0.2m}{s} \\ So, acceleration = \frac{Force}{Mass}= \frac{5(g+0.2)}{5} = \frac{(g+0.2m)}{s^2} 28 Let m_1 = 1 kg, m_2 = 2 kg and m_3 = 3 kg in figure (5-E12). Find the accelerations of m_1 , m_2 and m_3. The string from the upper pulley to m, is 20 cm when the system is released from rest. How long will it take before m, strikes the pulley ? Let the block m1+1 moves upward with acceleration a, and the two blocks m_2 and m_3 have relative acceleration a_2 due to the difference of weight between them. So, the actual acceleration at the blocks m_1,m_2,m_3 will be a_1.\\ (a_1-a_2) and (a_1+a_2) as shown \\ T = 1g-1a_2=0....(i) from fig(2) \\ \frac{T}{2}-2g-2(a_1-a_2)=0....(ii) from fig(3) \\ \frac{T}{2}-3g-3(a_1+a_2)=0....(iii) from fig(4) \\ From eqn (i) and (ii), eliminating T we get, 1g+1a_2=4g+4(a_1+a_2)\Rightarrow5a_2-4a_1=3g (iv) \\ From eqn (ii) and (iii), we get 2g + (a_1-a_2)=3g-3(a_1-a_2)\Rightarrow5a_1+a_2=(v) \\ Solving (iv) and (v), a_1 = \frac{2g}{29} and a_2 = g-5a_1=g-\frac{10g}{29}=\frac{19g}{29} \\ So, a_1 - a_2 = \frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29} \\ a_1 + a_2 =\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29} \\. So acceleration of m_1,m_2,m_3 are \frac{19g}{29}(up) \\ \frac{17g}{29} (down) \\ \frac{21g}{29}(down) respectively\\ Again, m_1,u=0,s=20cm=0.2m and a_2=\frac{19}{29}g [g=\frac{10m}{s^2}] \\ \therefore S =ut+\frac{1}{2}at^2=0.2=\frac{1}{2}\times\frac{19}{29}gt^2\Rightarrow t=0.25 sec. ##### Solution : Let the block m1+1 moves upward with acceleration a, and the two blocks m_2 and m_3 have relative acceleration a_2 due to the difference of weight between them. So, the actual acceleration at the blocks m_1,m_2,m_3 will be a_1.\\ (a_1-a_2) and (a_1+a_2) as shown \\ T = 1g-1a_2=0....(i) from fig(2) \\ \frac{T}{2}-2g-2(a_1-a_2)=0....(ii) from fig(3) \\ \frac{T}{2}-3g-3(a_1+a_2)=0....(iii) from fig(4) \\ From eqn (i) and (ii), eliminating T we get, 1g+1a_2=4g+4(a_1+a_2)\Rightarrow5a_2-4a_1=3g (iv) \\ From eqn (ii) and (iii), we get 2g + (a_1-a_2)=3g-3(a_1-a_2)\Rightarrow5a_1+a_2=(v) \\ Solving (iv) and (v), a_1 = \frac{2g}{29} and a_2 = g-5a_1=g-\frac{10g}{29}=\frac{19g}{29} \\ So, a_1 - a_2 = \frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29} \\ a_1 + a_2 =\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29} \\. So acceleration of m_1,m_2,m_3 are \frac{19g}{29}(up) \\ \frac{17g}{29} (down) \\ \frac{21g}{29}(down) respectively\\ Again, m_1,u=0,s=20cm=0.2m and a_2=\frac{19}{29}g [g=\frac{10m}{s^2}] \\ \therefore S =ut+\frac{1}{2}at^2=0.2=\frac{1}{2}\times\frac{19}{29}gt^2\Rightarrow t=0.25 sec. 29 In the previous problem, suppose m_2 = 2.0 kg and m_3 = 3.0 kg. What should be the mass m so that it remains at rest ? ##### Solution : m1 should be at rest \\ T-m_1{g}=0 \Rightarrow T=m_1g.....(i) \frac{T}{2}-2g-2a_1=0 \Rightarrow T-4g-4a_1=0.....(ii) \frac{T}{2}-3g-3a_1=0\Rightarrow T=6g-6a_1.....(iii) \\ From eqn(ii) & (iii) we get 3T-12g=12g-2T \Rightarrow T=\frac{24g}{5} = 408g \\ Putting the value of T eqn(i) we get m1 = 4.8kg. 30 Calculate the tension in the string shown in figure (5-E13). The pulley and the string are light and all surfaces are frictionless. Take g =\frac{10m}{s^2}. ##### Solution : T + 1a = 1g....(i) \\ T - 1a = 0 \Rightarrow T=1a....(ii) \\ From eqn (i) and (ii),we get \\ 1a+1a=1g \Rightarrow 2a=g\Rightarrow a=\frac{g}{2}=\frac{10}{2}=\frac{5m}{s^2} \\ From (ii) T = 1a=5N 31 Consider the situation shown in figure (5-E14). Both the pulleys and the string are light and all the surfaces are friction less. (a) Find the acceleration of the mass M. (b) Find the tension in the string. (c) Calculate the force exerted by the clamp on the pulley A in the figure. ##### Solution : Ma-2T=0 \Rightarrow Ma=2T \Rightarrow T=\frac{Ma}{2}.\\ T+Ma-Mg=0 \Rightarrow \frac{Ma}{2}+ma=Mg (because T=\frac{Ma}{2}) \\ \Rightarrow 3Ma =2 Mg\Rightarrow a=\frac{2g}{3} \\ a) acceleration of mass M is \frac{2g}{3} b) Tension T = \frac{Ma}{2}= \frac{M}{2}=\frac{2g}{3}=\frac{Mg}{3} \\ c) Let, R^1 = resultant of tensions = force exerted by the clamp on the pulley \\ R^1=\sqrt{T^2+T^2}=\sqrt{2T} \\ \therefore = \sqrt{2T}=\sqrt{2}\frac{Mg}{3}= \frac{\sqrt{2}{Mg}}{3} \\ Again, Tan\theta = \frac{T}{T}=1 \Rightarrow \theta = 45^o \\ So, it is \frac{\sqrt{2}{Mg}}{3} at an angle of 45^o with horizontal. \\ 32 Find the acceleration of the block of mass M in the situation shown in figure (5-E15). All the surfaces are frictionless and the pulleys and the string are light. ##### Solution : 2Ma + Mg sin\theta-T=0$$\\$ $\Rightarrow T=2Ma +Mg$ $sin\theta....(i)$ $\\$ $2T +2Ma -2Mg = 0$ $\\$ $\Rightarrow 2(2Ma +Mg$ $sin\theta)+2Ma-2Mg=0 [From (i)]$ $\\$ $\Rightarrow 4Ma+2Mg$ $sin\theta+2Ma-2Mg=0$ $\\$ $\Rightarrow 6Ma+2Mg$ $sin30^o-2Mg=0$ $\\$ $\Rightarrow 6Ma=Mg\rightarrow a=\frac{g}{6}$ $\\$ Acceleration of mass $M$ is $2a=2 \times \frac{g}{6}= \frac{g}{3}$ up the plane

33   Find the mass $M$ of the hanging block in figure (5-E16) which will prevent the smaller block from slipping over the triangular block. All the surfaces are friction less and the strings and the pulleys are light.

##### Solution :

As the block $m$ does not slinover $M^l$, ct will have acceleration as that of $M^l$ $\\$ From the freebody diagrams.$\\$ $T+Ma-Mg=0....(i)$ $\\$ $T-M'a-R$ $sin$ $\theta = 0....(ii)$ $\\$ $R$ $sin$ $\theta -ma = 0....(iii)$ $\\$ $R$ $cos$ $\theta -mg = 0....(iv)$ $\\$ Eliminating $T$, $R$ and $a$ from the above equation,we get $M$ =$\frac{M^l+m}{cot\theta-1}$ $\\$

34   Find the acceleration of the blocks A and B in the three situations shown in figure (5-E17).

##### Solution :

$a)$ $5a+T-5g=0 \Rightarrow T=5G-5a.....(i)$ $\\$ Again $(\frac{1}{2}-4g-8a=0\Rightarrow T=8g-16a.....(ii)$ $\\$ From eqn$(i)$ and $(ii)$, we get $5g-5a=8g+16a \Rightarrow 21a = -3g\Rightarrow a = - \frac{1}{7}g$ $\\$ So, the acceleration of $5kg$ mass is $\frac{g}{7}$ upward and that of $4 kg$ mass is $2a = \frac{2g}{7}$ (downward). $\\$

$b)$ $4a-V2=0\Rightarrow 8a-T=0\Rightarrow T = 8a.....(ii)$ $\\$ Again, $T + 5a - 5g=0\Rightarrow 8a+5a+5g=0$ $\\$ $\Rightarrow 13a-5g=0 \Rightarrow a=\frac{5g}{13}$ downward. $\\$ Acceleration of mass$(A)$ $kg$ is $2a$ = $\frac{10}{13} (g)$ & $5kg$ $(B)$ is $\frac{5g}{13}$ $\\$

$c)$ $T+1a-1g=0\Rightarrow T=1g-1a.....(i)$ $\\$ Again, $\frac{T}{2}-2g-4a=0\Rightarrow T-4g-8a=0.....(ii)$ $\\$ $\Rightarrow 1g-1a-4g-8a = 0$ $\\$ $[From (i)]$ $\\$ $\Rightarrow a= -(\frac{g}{3}) downward$ $\\$ Acceleration of mass $1kg(b)$ is $\frac{g}{3}(up)$ $\\$ Acceleration of mass $2kg(A)$ is $\frac{2g}{3}(downward)$ $\\$

35   Find the acceleration of the $500 g$ block in figure (5-E18).

##### Solution :

$m1 = 100g = 0.1kg$ $\\$ $m2 = 500g = 0.5kg$ $\\$ $m3 = 50g = 0.05kg$ $\\$ $T + 0.5a - 0.5g=0.....(i)$ $\\$ $T_1 - 0.5a - 0.05g=a.....(ii)$ $\\$ $T_1 + 0.1a -T+ 0.05g=0.....(iii)$ $\\$ From eqn $(ii) T_1=0.05g+0.05a.....(iv)$ $\\$ From eqn $(i) T_1=0.5g-0.5a.....(v)$ $\\$ Eqn $(ii)$ becomes $T_1+0.1a-T+0.05g=0$ $\\$ $\Rightarrow 0.05g+0.05a+0.1a-0.5g+0.5a+0.05g=0$ $[From(iv) and (v)]$ $\\$ $\Rightarrow 0.65a=0.4g \Rightarrow a =\frac{0.4}{0.65}=\frac{40}{65}g=\frac{8}{13}g downward.$ $\\$ Acceleration of $500gm$ block is $\frac{8g}{13g}$ downward.

36   A monkey of mass $15$ $kg$ is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of $\frac{1m}{s^2}$, how much force should it apply to the rope ? If the rope is $5$ $m$ long and the monkey starts from rest, how much time will it take to reach the ceiling ?

##### Solution :

$m$ = $15 kg$ of monkey. $\\$ $a=\frac{1m}{s^2}$ $\\$ From the free body diagram $\\$ $\therefore$ $T-[15g+15(1)]=0\Rightarrow T=15(10+1)\Rightarrow T=15 \times 11 \Rightarrow T=165N$ $\\$ The monkey should apply $165N$ force to the rope.$\\$ Initial velocity $u$=$0$; acceleration $a=\frac{1m}{s^2};s=5m$. $\\$ $\therefore s=ut+\frac{1}{2}at^2$ $\\$ $5=0+(\frac{1}{2})1 t^2$ $\\$ $\Rightarrow t^2=5 \times 2$ $\\$ $\Rightarrow t=\sqrt{10}$sec $\\$ TIme required is $\sqrt{10} sec$

37   A block $A$ can slide on a frictionless incline of angle $\theta$ and length $1$, kept inside an elevator going up with uniform velocity $v$ (figure 5-E22). Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.

##### Solution :

The driving force on the block which n the body to move down the plane is $F$ =$mg$ $sin$ $\theta$ . $\\$ So , acceleration = $g$ $sin$ $\theta$ $\\$ Initial velocity of block $u$ = $0$ $\\$ $s=l, a=$$mg sin \theta \\ Now, S = ut+\frac{1}{2}at^2 \\ \rightarrow l=0+\frac{1}{2l} (g sin \theta) t^2 \Rightarrow g^2=\frac{2l}{g sin \theta} \Rightarrow t=$$\sqrt\frac{2l}{g sin \theta}$ $\\$ Time taken is $\sqrt\frac{2l}{g sin \theta}$ $\\$

38   A monkey is climbing on a rope that goes over a smooth light pulley and supports a block of equal mass at the other end (figure 5-E19). Show that whatever force the monkey exerts on the rope, the monkey and the block move in the same direction with equal acceleration. If initially both were at rest, their separation will not change as time passes.

Suppose the monkey accelerates upward with acceleration $'a'$ & the block, accelerate downward with acceleration $a_1$. let the force exerted by the monkey is equal to $'T'$ $\\$ From the free body diagram of the monkey $\\$ $\therefore -mg-ma=0$ $....(i)$$\\ \Rightarrow T=mg+ma \\ Again from the FBD of the block, \\ T=ma_1 -mg=0 \\ \Rightarrow mg+ma+ma_1-mg=0 [From (i)]\Rightarrow ma-ma_1\Rightarrow a=a_1 \\ Acceleration '-a' downward i.e 'a' upward. \therefore The block & the monkey move in the same direction with equal acceleration. \\ If initially they are rest (no force is exerted by monkey)no motion of monkey of block occurs as they have same weight (same mass). their seperation will not change as time passes. 39 The monkey B shown in figure (5-E20) is holding on to the tail of the monkey A which is climbing up a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it ? Take g = \frac{10m}{s^2}. ##### Solution : Suppose 'A' move upward with acceleration a, such that in the tail of A maximum tension 30N produced. \\ T-5g-30-5a=0.....(i) \\ 30-2g-2a=0 ......(ii) \\ \Rightarrow T=5+30+(5 \times 5)=105N (max) \\ \Rightarrow 30-20-2a=0\Rightarrow a=\frac{5m}{s^2} \\ So. A can apply a maximum force of 105n in the rope to carry the monkey B with it. \\ For minimum force there is no acceleration of mnkey 'A' and B \Rightarrow a=0 \\ Now equation (ii) is T^l_1-2g=0\Rightarrow T^l_1=20N (wt of monkey B) \\ Equation (ii) is T-5g-20=0 [As T^l_1 = 20N] \\ \Rightarrow T=5g+20=50+20=70N. \\ \therefore the monkey A should apply force between 70N and 105N to carry the monkey B with it. 40 A car is speeding up on a horizontal road with an acceleration a. Consider the following situations in the car. (i) A ball is suspended from the ceiling through a string and is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal. ##### Solution : From the free body diagram T cos \theta-mg=0 \\ \Rightarrow T cos \theta = mg \\ T=\frac{mg}{cos\theta}.....(i) \\ ma-T$$sin$ $\theta$=$0$ $\\$ $\Rightarrow t=\frac{ma}{sin\theta}.....(ii)$ $\\$ From $(i)$ & $(ii)$ $\frac{mg}{cos\theta}=\frac{ma}{sin\theta}\Rightarrow$ $tan$ $\theta$ = $\frac{a}{g} \Rightarrow \theta$=$tan^{-t}$$\frac{a}{g} \\ The angle is Tan^{-t} \frac{a}{g} with vertical \\ (ii)m \Rightarrow mass of block \\ Suppose the angle of incline is \theta \\ From the diagram \\ ma cos \theta-mg sin \theta = 0 \\ \Rightarrow ma cos \theta =mg sin \theta \\ \Rightarrow \frac{sin \theta}{cos\theta} = \frac{a}{g} \\ \Rightarrow tan \theta = \frac{a}{g}\Rightarrow \theta = tan^-1(\frac{a}{g}) \\ Suppose pendulum makes \theta angle with the vertical. let, m = mass of the pendulum. \\ From the free body diagram T cos \theta-mg=0 \\ \Rightarrow T cos \theta = mg \\ T=\frac{mg}{cos\theta}.....(i) \\ ma-T$$sin$ $\theta$=$0$ $\\$ $\Rightarrow t=\frac{ma}{sin\theta}.....(ii)$ $\\$

From $(i)$ & $(ii)$ $\frac{mg}{cos\theta}=\frac{ma}{sin\theta}\Rightarrow$ $tan$ $\theta$ = $\frac{a}{g} \Rightarrow \theta$=$tan^{-t}$$\frac{a}{g}$ $\\$ The angle is $Tan^{-t}$ $\frac{a}{g}$ with vertical $\\$ $(ii)m \Rightarrow$ mass of block $\\$ Suppose the angle of incline is $\theta$ $\\$ From the diagram $\\$ $ma$ $cos$ $\theta$-$mg$ $sin$ $\theta$ = $0$ $\\$ $\Rightarrow$ $ma$ $cos$ $\theta$ =$mg$ $sin$ $\theta$ $\\$ $\Rightarrow$ $\frac{sin \theta}{cos\theta}$ = $\frac{a}{g}$ $\\$ $\Rightarrow$ $tan$ $\theta$ = $\frac{a}{g}\Rightarrow \theta = tan^-1(\frac{a}{g})$ $\\$

41   6. A $3 cm$ tall object is placed at a distance of $7.5 cm$ from a convex mirror of focal length $6 cm.$ Find the location, size and nature of the image.

##### Solution :

42   A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of $\frac{12m}{s^2}$. Find the displacement of the block during the first $0.2 s$ after the start. Take $g = \frac{10m}{s^2}$.

##### Solution :

Because, the elevator is moving downward with an acceleration $\frac{12m}{s^2}$ $(>g)$, the body gets seperated. So, body moves with an acceleration $g=\frac{10m}{s^2}$ [freely falling body] and the elevator move with acceleration $\frac{12m}{s^2}$ $\\$ Now, the block has acceleration $g$=$\frac{10}{ms^2}$ $\\$ $u$ = $0$ $\\$ $t$ = $0.2 sec$ $\\$ So, the distance traveled by the block is given by, $\\$ $\therefore s=ut+\frac{1}{2}at^2$ $\\$ $=0+(\frac{1}{2})10(0.2)^2=5 \times 0.04 = 0.2m = 20cm$. $\\$ The displacement of body is $20cm$ during first $0.2 sec$

43   Figure (5-E21) shows a man of mass $60$ $kg$ standing on a light weighing machine kept in a box of mass $30$ $kg$. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, what is the weight shown by the machine ? What force should he exert on the rope to get his correct weight on the machine?

##### Solution :

$(i)$ Given , Mass of man = $60kg$ $\\$ Let $R^l$ = apparent weight of man in this case . $\\$ Now, $R^l +T -60g=0$ $[From FBD of man]$ $\\$ $\Rightarrow T=60g-R^l.....(i)$ $\\$ $T-R^l-30g=0.....(ii)$ $[From FBD of box]$ $\\$ $\Rightarrow 60g-R^l-R^l-30g=0$ $[From (i)]$ $\\$ $\Rightarrow R^l=15g$ The weight shown by the machine is 15kg.$\\$

$(ii)$ To get his correct weight suppose the applied force is $T$ and so, accelerates upward with $'a'$. $\\$ In this case, given that correct weight$=R=60g$, where $g=acc^n$ due to gravity $\\$ From the FBD of the man $T^1+R-60g-60a=0$ $\\$ $T^1-60a=0[\therefore R=60g]$ $T^1=60a.....(i)$ $\\$ From the FBD of the box $\\$ $T^1-R-30g-30a=0$ $\\$ $\Rightarrow T^1-60g-30g-30a=0$ $\\$ $\Rightarrow T^1-30a=90g=900$ $\\$ $\Rightarrow T^1=30a-900 .....(ii)$ $\\$ From the eq $(i)$ and$(ii)$ we get $T^1=2T^1-1800\Rightarrow T^1=1800N$ $\\$ $\therefore$ So, he should exert $1800N$ force on the rope to get correct reading.