**1.** The mass of cyclist together with the bike is $90$ kg.
Calculate the increase in kinetic energy if the speed
increases from $6'0$ km/h to $12$ km/h.

$M$ = $m_c$ + $m_b$ = $90kg$ $\\$ $u$ = $6 km/h$ = $1.666 m/sec$ $\\$ $v$ = $12 km/h$ = $3.333$ m/sec $\\$ $increase$ in $K.E.$ = $\frac{1}{2}$ $Mv^2$ - $\frac{1}{2}$ $Mu^2$ $\\$ = $\frac{1}{2}$ $90$ $\times$ $(3.333)^2$ - $\frac{1}{2}$ $\times$ $90$ $\times$ $(1.66)^2$ = $494.5$ - $124.6$ = $374.8$ $\approx$ $375J$

$increase$ in $K.E.$ = $\frac{1}{2}$ $Mv^2$

**2.** A block of mass $2.00$ kg moving at a speed of $10.0$ m/s
accelerates at $3.00$ m/s for $5.00$ s. Compute its final
kinetic energy.

$m_b$ = $2kg$ $\\$ $u$ = $10 m/sec$ $\\$ $a$ = $3 m/aec^2$ $\\$ $t$ = $5 sec$ $\\$ $v$ = $u$ + $at$ = $10$ + $3 I 5$ = $25 m/sec$ $\\$ $\therefore$ $F.K.E$ = $\frac{1}{2}$ $mv^2$ = $\frac{1}{2}$ $\times$ $2$ $\times$ $625$ = $625 J$

**3.** A box is pushed through $4.0$ m across a floor offering
$100$ N resistance. How much work is done by the
resisting force ?

$F$ = $100 N$ $\\$ $S$ = $4m,$ $\theta$ = $0^o$ $\\$ $\omega$ = $F.S$ = $100$ $\times$ $4$ = $400$ $J$

**4.** A block of mass $5.0$ kg slides down an incline of
inclination $30°$ and length $10$ m. Find the work done by
the force of gravity.

$m$ = $5kg$ $\\$ $\theta$ = $30^o$ $\\$ $S$ = $10$ $m$ $F$ = $mg$ $\\$ $ So$ $work$ $done$ $by$ $the$ $force$ $of$ $gravity$ $\\$ $\omega$ = $mgh$ = $5$ $\times$ $9.8$ $\times$ $5$ = $245$ $J$

**5.** A constant force of $2.50$ N accelerates a stationary
particle of mass $15$ g through a displacement of $2.50$ m.
Find the work done and the average power delivered.

$F$ = $2.50N$ , $S$ = $2.5m,$ $m$ =$15g$ = $0.015kg$ $\\$ $So,$ $w$= $F$ $\times$ $S$ $\Rightarrow$ $a$ = $\frac{F}{m}$ = $\frac{2.5}{0.015}$ = $\frac{500}{3}$ $m/s^2$ $\\$ = $F$ $times$ $S$ $cos$ $0^o$ $($ $acting$ $alone$ $the$ $same$ $line$$)$ $\\$ $2.5$ $\times$ $2.5$ = $6.25$ $J$ $\\$ $Let$ $the$ $velocity$ $of$ $the$ $body$ $at$ $b$ = $U.$ $Applying$ $work$ $energy$ $principle$ $\frac{1}{2}$ $mv^2$ - $0$ = $6.25$ $\\$ $\Rightarrow$ $V$ = $\sqrt{\frac{6.25 \times 2}{0.015}}$ = $28.86$ $m/sec$ $\\$ $So$ $time$ $taken$ $to$ $travel$ $from$ $A$ $to$ $B.$ $\\$ $\Rightarrow$ $t$ = $\frac{v-u}{a}$ = $\frac{28.86 \times 3}{500}$ $\\$ $\therefore$ $Average$ $Power$ = $\frac{W}{t}$ = $\frac{6.25 \times 500}{(28.86) \times 3}$ = $36.1$

**6.** A particle moves from a point $\vec{r_1}$ = (2 m)$\vec{i}$ + (3 m)$\vec{j}$ to
another point $\vec{r_2}$ = (3 m)$\vec{i}$ + (2 m)$\vec{j}$ during which a certain
force $\vec{F}$ = (5 N)$\vec{i}$ + (5 N)$\vec{j}$ acts on it. Find the work done
by the force on the particle during the displacement.

$\vec{r_1}$ = $2\hat{i}$ + $3\hat{j}$ $\\$ $r_2$ = $3\hat{i}$ + $2\hat{j}$ $\\$ $So,$ $displacement$ $vector$ $is$ $given$ $by,$ $\\$ $\vec{r}$ = $\vec{r_1}$ - $\vec{r_2}$ $\Rightarrow$ $\vec{r}$ = $(3\hat{i} + 2\hat{j})$ - $(2\hat{i} + 3\hat{j})$ = $\hat{i} - \hat{j}$ $\\$ $So,$ $Work$ $done $ = $\vec{F}$ $\times$ $\vec{s}$ = $5 \times 1 + 5(-1) = 0$

**7.** A man moves on a straight horizontal road with a block
of mass $2$ kg in his hand. If he covers a distance of $40$ in
with an acceleration of $0.5$ $m/s^2$, find the work done by
the man on the block during the motion.

$m_b = 2kg,$ $s = 40m,$ $a = 0.5m/sec^2$ $\\$ $So, force applied by the man on the box$ $\\$ $F = m_b a = 2 \times (0.5) = 1 N$ $\\$ $\omega = FS = 1 \times 40 = 40J$

**8.** A force $F = a + bx$ acts on a particle in the $x$-direction,
where $a$ and $b$ are constants. Find the work done by this
force during a displacement from $x = 0$ $to$ $x = d.$

8 None

SolutionsGiven That $F = a + bx$ $\\$ Where $a$ and $b$ are constants. $\\$ So, work done by this force during this force during the displacement $x = 0$ and x = d is given by $\\$ W = $\int_0^d F\; \mathrm{d}x$ = $\int_0^d (a + bx)\; \mathrm{d}x$ = $ax + (bx^2/2 )$ = $[a + \frac{1}{2} bd] d$

**9.** A block of mass $250$ g slides down an incline of
inclination $37°$ with a uniform speed. Find the work done
against the friction as the block slides through $1.0$ m.

$m_b = 250 g = .250 kg$ $\\$ $\theta = 37^o, S = 1m.$ $\\$ Frictional force f = $\mu$ R $\\$ $\\$ mg sin $\theta$ = $\mu$ R $\quad$ ...(1) $\\$ mg cos $\theta$ $\qquad$ $\ $ ...(2) $\\$ So, work done against $\mu$R = $\mu$RS cos $0^o$ = mg sin $\theta$ S = 0.250 $\times$ 9.8 $\times$ 0.60 $\times$ 1 = 1.5 J

**10.** A block of mass $m$ is kept over another block of mass
$M$ and the system rests on a horizontal surface
(figure 8-E1). A constant horizontal force $F$ acting on the
lower block produces an acceleration $\frac{F}{2(m + M)}$
in the
system, the two blocks always move together. (a) Find
the coefficient of kinetic friction between the bigger block
and the horizontal surface. (b) Find the frictional force
acting on the smaller block. (c) Find the work done by
the force of friction on the smaller block by the bigger
block during a displacement $d$ of the system.

a = $\frac{F}{2(m + M)}$ (given) $\\$ a) from fig (1) $\\$ ma = $\mu_k$ $R_1$ and $R_1$ = mg $\\$ $\Rightarrow$ $\mu$ = $\frac{ma}{R_1}$ = $\frac{F}{2(m + M)g}$ $\\$ b) Frictional force acting on the smaller block f = $\mu$ R = $\frac{F}{2(m + M)g} \times mg = \frac{m \times F}{2(M + m)}$ $\\$ c) work done w = fs $\qquad$ s = d $\\$ w = $\frac{mF}{2(M + m)} \times d = \frac{mFd}{2(M + m)}$

**11.** A box weighing $2000$ N is to be slowly slid through $20$ m
on a straight track having friction coefficient $0.2$ with
the box. (a) Find the work done by the person pulling
the box with a chain at an angle $\theta$ with the horizontal.
(b) Find the work when the person has chosen a value
of $\theta$ which ensures him the minimum magnitUde of the
force.

Weight = 2000 N, S = 20m, $\mu$ = 0.2 $\\$ a) R + P sin $\theta$ - 2000 = 0 $\quad$ ..(1) $\\$ P cos $\theta$ - 0.2 R = 0 $\qquad$ ...(2) $\\$ From (1) and (2) P cos $\theta$ - 0.2 (2000 - P sin $\theta$) = 0 $\\$ P = $\frac{400}{cos \theta + 0.2 sin \theta}$ $\qquad$ ...(3) $\\$ So, work done by the person , W = PS cos $\theta$ = $\frac{8000 cos \theta }{cos \theta + 0.2 sin \theta}$ = $\frac{8000}{1 + 0.2 sin \theta}$ = $\frac{40000}{5 + tan \theta}$ $\\$ b) for minimum magnitude of force from equn (1) $\\$ $\frac{d}{d \theta}$ (cos $\theta$ + 0.2 sin $\theta$) = 0 $\Rightarrow$ tan $\theta$ = 0.2 $\\$ putting the value in equn (3) $\\$ W = $\frac{40000}{5 + tan \theta} = \frac{40000}{(5.2)} = 7690 J$

**12.** A block of weight $100$ N is slowly slid up on a smooth
incline of inclination $37°$ by a person. Calculate the work
done by the person in moving the block through a
distance of $2.0$ m, if the driving force is (a) parallel to
the incline and (b) in the horizontal direction.

w = 100 N, $\theta$ = $37^o$, s = 2m $\\$ Force F = mg sin $37^o$ = 100 $\times$ 0.60 = 60 N $\\$ So, work done, when the force is parallel to incline. $\\$ w = Fs cos $\theta$ = 60 $\times$ 2 $\times$ cos $\theta$ = 120 J $\\$ In $\Delta$ABC AB = 2m $\\$ CB = $37^o$ $\\$ So, h = c = 1m $\\$ $\therefore$ work done when the force in horizontal direction $\\$ W = mgh = 100 $\times$ 1.2 = 120 J

**13.** Find the average frictional force needed to stop a car
weighing $500$ kg in a distance of $25$ m if the initial speed
is $72$ km/h.

m = 500 kg , s = 25m, u = 72 km/h = 20 m/s, $\\$ (-a) = $\frac{v^2 - u^2}{2S} \Rightarrow a = \frac{400}{50} = 8 m/sec^2$ $\\$ Frictional force f = ma = 500 $\times$ 8 = 4000 N

**14.** Find the average force needed to accelerate a car
weighing $500$ kg from rest to $72$ km/h in a distance of
$25$ m.

m = 500 kg, u = 0, v = 72 km/h = 20 m/s $\\$ a = $\frac{v^2 - u^2}{2s} = \frac{400}{50} = 8 m/sec^2$ $\\$ force needed to accelerate the car F = ma = 500 $\times$ 8 = 4000 N

**15.** A particle of mass $m$ moves on a straight line with its
velocity varying with the distance travelled according to
the equation $v = a \sqrt x,$ where $a$ is a constant. Find the
total work done by all the forces during a displacement
from $x = 0$ to $x = d$

Given $v = a \sqrt x,$ (uniformly accelerated motion) $\\$ displacement s = d - 0 = d $\\$ putting x = 0, $\qquad$ $v_1$ = 0 $\\$ putting x = d, $\qquad$ $v_2$ = a $\sqrt d$ $\\$ a = $\frac{v_2^2 - u_2^2}{2s} = \frac{a^2 d}{2d} = \frac{a^2}{2}$ $\\$ force f = ma = $\frac{ma^2}{2}$ $\\$ work done w = FS cos $\theta$ = $\frac{ma^2}{2} \times d = \frac{ma^2 d}{2}$

**16.** A block of mass $2.0$ kg kept at rest on an inclined plane
of inclination $37°$ is pulled up the plane by applying a
constant force of $20$ N parallel to the incline. The force
acts for one second. (a) Show that the work done by the
applied force does not exceed $40$ J. (b) Find the work
done by the force of gravity in that one second if the
work done by the applied force is $40$ J. (c) Find the
kinetic energy of the block at the instant the force ceases
to act. Take $g = 10 m/s^2.$

m = 2 kg, $\theta$ = $37^o$, F = 20 N $\\$ From the free body diagram $\\$ F = (2g sin $\theta$) + ma $\Rightarrow$ a = (20 - 20 sin $\theta$ )/s = 4 $m/sec^2$ $\\$ S = ut + $\frac{1}{2} at^2$ $\qquad$ (u=0, t=1s, a = 1.66) = 2m $\\$ So, work done w = Fs = 20 $\times$ 2 = 40 J $\\$ b) If W = 40 J $\\$ S = $\frac{W}{F} = \frac{40}{20}$ $\\$ h = 2 sin $37^o$ = 1.2 m $\\$ So, work done W = -mgh = -20 $\times$ 1.2 = -24 J $\\$ c) v = u + at = 4 $\times$ 10 = 40 m/sec $\\$ So, K.E. = $\frac{1}{2} mv^2 = \frac{1}{2} \times 2 \times 16 = 16 J $

**17.** A block of mass $2.0$ kg is pushed down an inclined plane
of inclination $37°$ with a force of $20$ N acting parallel to
the incline. It is found that the block moves on the
incline with an acceleration of $10$ $m/s^2.$ If the block started from rest, find the work done (a) by the applied
force in the first second, (b) by the weight of the block
in the first second and (c) by the frictional force acting
on the block in the first second. Take $g$ = $10$ $m/s^2.$

17 None

Solutionsm = 2 kg, $\theta$ = $37^o$, F = 20 N, a = 10 $m/sec^2$ $\\$ a) t = 1sec $\\$ So, s = ut + $\frac{1}{2} at^2$ = 5m $\\$ work done by the applied force w = FS cos $0^o$ = 20 $\times$ 5 = 100 J $\\$ b) BC (h) 5 sin $37^o$ = 3m $\\$ So, work done by the weight W = mgh = 2 $\times$ 10 $\times$ 3 = 60 J $\\$ c) So, frictional force f = mg sin $\theta$ $\\$ work done by the frictional forces w = fs cos $0^o$ = (mg sin $\theta$) s = 20 $\times$0.60 $\times$ 5 = 60 J

**18.** A $250$ g block slides on a rough horizontal table. Find
the work done by the frictional force in bringing the
block to rest if it is initially moving at a speed of $40$
cm/s. If the friction coefficient between the table and the
block is $0.1,$ how far does the block move before coming
to rest ?

Given m = 250 g = 0.250 kg, $\\$ u = 40 cm/sec = 0.4 m/sec $\\$ $\mu$ = 0.1 , v = 0 $\\$ Here, $\mu$ R = ma {where a = deceleration} $\\$ a = $\frac{\mu R}{m} = \frac{\mu mg}{m} = \mu g = 0.1 \times 9.8 = 0.98 m/sec^2$ $\\$ S = $\frac{v^2 - u^2}{2a} = 0.082m = 8.2 cm$ $\\$ Again, work done against friction is given by $\\$ - w = $\mu$ RS cos $\theta$ $\\$ = 0.1 $\times$ 2.5 $\times$ 0.082 $\times$ 1($\theta = 0^o$ ) = 0.02 J $\\$ $\Rightarrow$ W = -0.02 J

**19.** Water falling from a $50$ m high fall is to be used for
generating electric energy. If $1.8 \times 10^5$ kg of water falls
per hour and half the gravitational potential energy can
be converted into electric energy, how many $100$ W
lamps can be lit ?

h = 50 m, m = $1.8 \times 10^5$ kg/hr, P = 100 watt, $\\$ P.E. = mgh = $1.8 \times 10^5 \times 9.8 \times 50 = 882 \times 10^5 J/hr$ $\\$ Because, half the potential energy is converted into electricity, $\\$ Electrical energy $\frac{1}{2}$ P.E. = 441 $\times 10 ^5 J/hr$ $\\$ So, power in watt (J/sec) is given by = $\frac{441 \times 10^5}{3600}$ $\therefore$ number of 100 W lamps, that can be lit $\frac {441 \times 10^5}{3600 \times 100} = 122.5 \approx 122$

**20.** A person is painting his house walls. He stands on a
ladder with a bucket containing paint in one hand and
a brush in other. Suddenly the bucket slips from his
hand and falls down on the floor. If the bucket with the
paint had a mass of $6.0$ kg and was at a height of $2.0$ m
at the time it slipped, how much gravitational potential
energy is lost together with the paint ?

m = 6kg, h = 2m $\\$ P.E. at a height '2m' = mgh 6 $\times$ 9.8 $\times$ 2 = 117.6 J $\\$ P.E. at floor = 0 $\\$ Loss in P.E. = 117.6 - 0 = 117.6 J $\approx$ 118 J

**21.** A projectile is fired from the top of a $40$ m high cliff
with an initial speed of $50$ $m/s$ at an unknown angle.
Find its speed when it hits the ground.

h = 40m, u = 50 m/sec $\\$ Let the speed be 'v' when it strikes the ground. $\\$ Applying law of conservation of energy $\\$ mgh + $\frac{1}{2} mu^2 = \frac{1}{2} mv^2$ $\\$ $\Rightarrow 10 \times 40 +(\frac{1}{2}) \times 2500 = \frac{1}{2} v^2 \Rightarrow v^2 = 3300 \Rightarrow v = 57.4 m/sec \approx 58 m/sec$

**22.** The $200$ $m$ free style women's swimming gold medal at
Seol Olympic $1988$ went to Heike Friendrich of East
Germany when she set a new Olympic record of $1$ minute
and $57.56$ seconds. Assume that she covered most of the
distance with a uniform speed and had to exert $460$ W
to maintain her speed. Calculate the average force of
resistance offered by the water during the swim.

t = 1 min 57.56 sec = 11.56 sec, p = 400 w, s = 200m $\\$ p = $\frac{w}{t}$, work w = pt = $460 \times 117.56 J$ $\\$ Again W = FS = $\frac{460 \times 117.56}{200} = 270.3 \approx 270 N $

**23.** The US athlete Florence Griffith-Joyner won the $100$ m
sprint gold medal at Seol Olympic $1988$ setting a new
Olympic record of $10.54$ s. Assume that she achieved her
maximum speed in a very short-time and then ran the
race with that speed till she crossed the line. Take her
mass to be $50$ kg. (a) Calculate the kinetic energy of
Griffith-Joyner at her full speed. (b) Assuming that the
track, the wind etc. offered an average resistance of one
tenth of her weight, calculate the work done by the
resistance during the run. (c) What power Griffith-
Joyner had to exert to maintain uniform speed ?

S = 100 m, t = 10.54 sec, m = 50 kg $\\$ The motion can be assumed to be uniform because the time take for acceleration is minimum. $\\$ a) Speed v = $\frac{S}{t} = 9.487$ $e/s$ $\\$ So, K.E. = $\frac{1}{2} mv^2 = 2250J$ $\\$ b) Weight = mg = 490 J $\\$ given R = mg /10 = 49 J $\\$ So, work done against resistance $W_f = - RS = - 49 \times 100 = - 4900 J$ $\\$ c) To maintain her uniform speed , she has to expert 4900 j of energy to overcome friction $\\$ P = $\frac{W}{t} = 4900 / 10.54 = 465 W$

**24.** A water pump lifts water from a level $10$ $m$ below the
ground. Water is pumped at a rate of $30$ $kg/minute$ with
negligible velocity. Calculate the minimum horsepower
the engine should have to do this.

h = 10 m $\\$ flow rate = $(\frac{m}{t}) = 30 kg/min = 0.5 kg/sec $ $\\$ power P = $\frac{mgh}{t} = (0.5) \times 9.8 \times 10 = 49 W$ $\\$ So, horse power (h.p.) $P/746 = 49/746 = 6.6 \times 10^{-2} hp$

**25.** An unruly demonstrator lifts a stone of mass $200$ $g$ from
the ground and throws it at his opponent. At the time
of projection, the stone is $150$ $cm$ above the ground and
has a speed of $3.00$ $m/s.$ Calculate the work done by the
demonstrator during the process. If it takes one second
for the demonstrator to lift the stone and throw, what
horsepower does he use ?

m = 200 g = 0.2kg , h = 150 cm = 1.5 m, v = 3 m/sec, t = 1 sec $\\$ Total work done = $\frac{1}{2} mv^2 + mgh$ $\\$ $= (1/2) \times (0.2) \times 9 + (0.2) \times (9.8) \times (1.5) = 3.84 J$ $\\$ h.p. used = $\frac{3.84}{746} = 5.14 \times 10^{-3}$

**26.** In a factory it is desired to lift $2000$ kg of metal through
a distance of $12$ $m$ in $1$ minute. Find the minimum
horsepower of the engine to be used.

m = 200 kg, s = 12 m, t = 1 min = 60 sec $\\$ So, work W = F cos $\theta$ = mgs cos $0^o$ [$\theta = 0^o$ for minimum work] $\\$ $ = 2000 \times 10 \times 12 = 240000J $ $\\$ So, power p = $\frac{W}{t} = \frac{240000}{60} = 4000 watt $ $\\$ h.p. = $\frac{4000}{746} = 5.3 hp.$

**27.** A scooter company gives the following specifications
about its product. $\\$
Weight of the scooter - $95$ kg $\\$
Maximum speed - $60$ $km/h$ $\\$
Maximum engine power - $3.5$ $hp$ $\\$
Pick up time to get the maximum speed - $5$ $s$ $\\$
Check the validity of these specifications

The specification given by the company are $\\$ U = 0, m = 95 kg, $p_m = 3.5$ hp $\\$ $V_m = 60 km/h = 50/3$ $m/sec \qquad t_m = 5 sec$ $\\$ So, the maximum acceleration that can be produced is given by, $\\$ $a = \frac{(50/3)-0}{5} = \frac{10}{3}$ $\\$ So, the driving force is given by $\\$ F = ma = $95 \times \frac{10}{3} = \frac{950}{3} N$ $\\$ So, the velocity that can be attained by maximum h.p. while supplying $\frac{950}{3}$ will be $\\$ $v = \frac{p}{F} \Rightarrow v = \frac{3.5 \times 746 \times 5}{950} = 8.2 m/sec.$ $\\$ Because the scooter can reach a maximum of 8.2 m/sec while producing a force of 950/3 N, the specification given are some what over claimed.

**28.** A block of mass $30.0$ $kg$ is being brought down by a
chain. If the block acquires a speed of $40.0$ $cm/s$ in
dropping down $2.00$ $m,$ find the work done by the chain
during the process.

28 None

SolutionsGiven m = 30 kg, v = 40 cm/sec = 0.4 m/sec, s = 2m $\\$ From the free body diagram, the force given by the chain is, $\\$ $F = (ma-mg) = m(a-g) $ [where a = acceleration of the block] $\\$ $a = \frac{v_2 - u_2 }{2s} = \frac{0.16}{0.4} = 0.04 m/sec^2$ $\\$ So, work done W=Fs cos $\theta$ = m(a-g) s cos $\theta$ $\\$ $\Rightarrow W = 30 (0.04-9.8) \times 2 \Rightarrow W = -585.5 \Rightarrow W = -585.5 J.$ $\\$ So, $W = -586 J $

**29.** The heavier block in an Atwood machine has a mass
twice that of the lighter one. The tension in the string
is $16.0$ N when the system is set into motion. Find the
decrease in the gravitational potential energy during the
first second after the system is released from rest

Given T = 19 N $\\$ From the free body diagrams, $\\$ $T-2 mg + 2 ma = 0 \quad ...(1)$ $\\$ $T-mg - ma =0 \qquad ...(2)$ $\\$ From equation (1) $\&$ (2) T = 4ma $\Rightarrow a = \frac{T}{4m} \Rightarrow A = \frac{16}{4m} = \frac{4}{m} m/s^2$ $\\$ Now, S = $ut + \frac{1}{2} at^2$ $\\$ $\Rightarrow S = \frac{1}{2} \times \frac{4}{m} \times 1 \Rightarrow S = \frac{2}{m} m $ [because u= 0] $\\$ Net mass $=2m-m = m$ $\\$ Decrease in P.E. = mgh $\Rightarrow P.E. = m \times g \times \frac{2}{m} \Rightarrow P.E. = 9.8 \times2 \Rightarrow P.E. = 19.6 J$

**30.** The two blocks in an Atwood machine have masses
$2.0$ $kg$ and $3.0$ $kg.$ Find the work done by gravity during
the fourth second after the system is released from rest.

Given $m_1 = 3 kg, m_2 = 2kg \qquad$ t = during 4th second $\\$ From the free body diagram $\\$ $T-3g +3a = 0 \qquad ...(1)$ $\\$ $T-2g -2a = 0 \qquad ...(2)$ $\\$ Equation (1) $\&$ (2) we get $3g -3a = 2g + 2a \Rightarrow a = \frac{g}{5} m/sec^2$ $\\$ Distance traveled in 4th sec is given by $\\$ $S_{4th} = \frac{a}{2}(2n-1) = \frac{\frac{g}{5}}{s}(2 \times 4-1) = \frac{7g}{10} = \frac{7 \times 9.8}{10} m $ $\\$ Net mass 'm' $ = m_1- m_2= 3-2=1 kg$ $\\$ So, decrease in P.E.$ = mgh = 1 \times 9.8 \times \frac{7}{10} \times 9.8 = 67.2 =67 J $

**31.** Consider the situation shown in figure (8-E2). The system
is released from rest and the block of mass $1.0$ kg is
found to have a speed $0.3$ m/s after it has descended
through a distance of $1$ m. Find the coefficient of kinetic
friction between the block and the table.

$ m_1 = 4 kg, m_2 = 1 kg, V_2=0.3m/sec, V_1= 2\times (0.3) = 0.6 m/sec $ $\\$ $(v_1 = 2x_2$ m this system) $\\$ h = 1m = height decent by 1 kg block $\\$ $s = 2 \times 1 = 2 m$ distance traveled by 4 kg block $\\$ $u = 0$ $\\$ Applying change in K.E. = work done (for the system)$\\$ $[(1/2)m_1v_1^2 + (1/2)m_2v_m^2] - 0=(-\mu R)S + m_2g \qquad$ $\\$ [R = 4g = 40 N] $\\$ $\Rightarrow \frac{1}{2} \times 4 \times (0.36) \times \frac{1}{2} \times 1 \times (0.09) = - \mu \times 40 \times 2 + 1 \times 40 \times 1$ $\\$ $\Rightarrow 0.72 + 0.045 = - 80 \mu + 10$ $\\$ $\Rightarrow \mu \frac{9.235}{80} = 0.12$

**32.** A block of mass $100$ $g$ is moved with a speed of $5.0$ $m/s$
at the highest point in a closed circular tube of radius
$10$ $cm$ kept in a vertical plane. The cross-section of the
tube is such that the block just fits in it. The block
makes several oscillations inside the tube and finally
stops at the lowest point. Find the work done by the
tube on the block during the process.

Given, m = 100g = 0.1 kg, v = 5m/sec, r = 10 cm. $\\$ work done by the block = total energy at A - total energy at B $\\$ $(1/2 mv^2 + mgh ) - 0$ $\\$ $\Rightarrow W = \frac{1}{2} mv^2 + mgh -0 = \frac{1}{2} \times (0.1) \times 25 + (0.1) \times 10 \times (0.2)$ [h = 2r = 0.2 m] $\\$ $\Rightarrow W = 1.25-0.2 \Rightarrow W = 1.45 J $ $\\$ So, the work done by the tube on the body is $\\$ $W_t = -1.45 J$

**33.** A car weighing $1400$ $kg$ is moving at a speed of $54$ $km/h$
up a hill when the motor stops. If it is just able to reach
the destination which is at a height of $10$ $m$ above the
point, calculate the work done against friction (negative
of the work done by the friction).

m = 1400 kg , v = 54 km/h = 15 m/sec, h = 10 m $\\$ Work done = (total K.E.) - total P.E.$\\$ $= 0 + \frac{1}{2} mv^2 - mgh = \frac{1}{2} \times 1400 \times (15)^2 - 1400 \times 9.8 \times 10 \\ = 157500 - 137200 = 20300$ $\\$ So, work done against friction, $W_t = 20300 J$

**34.** A small block of mass $200$ $g$ is kept at the top of a
frictionless incline which is $10$ $m$ long and $3.2$ $m$ high.
How much work was required (a) to lift the block from
the ground and put it at the top, (b) to slide the block
up the incline ? What will be the speed of the block when
it reaches the ground, if (c) it falls off the incline and
drops vertically on the ground (d) it slides down the
incline ? Take $g$ = $10$ $m/s^2$

m = 200 g = 0.2 kg , s = 10 m, h = 3.2 m, g = 10 $m/sec^2$ $\\$ a) work done W = mgh = $0.2 \times 10 \times 3.2 = 6.4 J$ $\\$ b) work done to slide the block up the incline $\\$ $w = (mg \ sin \theta) = (0.2) \times 10 \times \frac{3.2}{10} \times 10 = 6.4 J$ $\\$ c) Let, the velocity be v when falls on the ground vertically, $\\$ $\frac{1}{2} mv^2 - 0 = 6.4 J \Rightarrow v = 8 m/s $ $\\$ d) Let V be the velocity when reaches the ground by liding $\\$ $\frac{1}{2} mV^2 - 0 = 6.4 J \Rightarrow v = 8 m/sec $ $\\$

**35.** In a children's park, there is a slide which has a total
length of $10$ $m$ and a height of $8.0$ $m$ (figure 8-E3).
Vertical ladder are provided to reach the top. A boy
weighing $200$ $N$ climbs up the ladder to the top of the
slide and slides down to the ground. The average friction offered by the slide is three tenth of his weight. Find
(a) the work done by the ladder on the boy as he goes
up, (b) the work done by the slide on the boy as he comes
down. Neglect any work done by forces inside the body
of the boy.

$l$ = 10m, h = 8m, mg = 200 N $\\$ $f = 200 \times \frac{3}{10} = 60 N$ $\\$ a) work done by the ladder on the boy is zero when the boy is going up because the work is done by the boy himself. $\\$ b) work done against frictional force, $W = \mu RS = f l = (-60) \times 10 = -600 J $ $\\$ c) work done by the force inside the boy is $\\$ $W_b = (mg\ sin \theta ) \times 10 = 200 \times \frac{8}{10} \times 10 = 1600 J$

**36.** 36. Figure (8-E4) shows a particle sliding on a frictionless
track which terminates in a straight horizontal section.
If the particle starts slipping from the point A, how far
away from the track will the particle hit the ground ?

H = 1m, h = 0.5m $\\$ Applying law of conservation of Energy for point A $\&$ B $\\$ $mgH = \frac{1}{2} mv^2+mgh \Rightarrow g = (1/2)v^2 + 0.5g \\ \Rightarrow v^2 2(g-0.59) = g \Rightarrow v= \sqrt g = 3.1 m/s$ $\\$ After point B the body exhibits projectile motion for which $\\$ $\theta = 0^o, v= -0.5 \\$ So, $-0.5 = (u\ sin \theta)t - (1/2) gt^2 \Rightarrow 0.5 = 4.9 t^2 \Rightarrow t = 0.31 sec \\$ So, $x = (4\ cos \theta)t = 3.1 \times 3.1 = 1m. \\$ So, the particle will hit the ground at a horizontal distance in from B.

**37.** A block weighing $10$ $N$ travels down a smooth curved
track $AB$ joined to a rough horizontal surface
(figure 8-E5). The rough surface has a friction coefficient
of $0.20$ with the block. If the block starts slipping on the
track from a point $1.0$ $m$ above the horizontal surface,
how far will it move on the rough surface ?

mg = 10N, $\mu = 0.2,$ H = 1m, u=v=0 $\\$ change in P.E. = work done. $\\$ Increase in K.E. $\\$ $\Rightarrow w = mgh = 10 \times 1 = 10 J \\$ Again on the horizontal surface the frictional force $\\$ $F = \mu R= \mu mg= 0.2 \times 10 = 2N \\$ So, the K.E. is used to overcome friction $\\$ $\Rightarrow S = \frac{W}{F} = \frac{10J}{2N}=5m$

**38.** A uniform chain of mass $m$ in and length $l$ overhangs a
table with its two third part on the table. Find the work
to be done by a person to put the hanging part back on
the table.

Let 'dx' be the length of an element at a distance x from the table $\\$ mass of 'dx' length = $(m/l)dx \\$ Work done to put dx part back on the table $\\$ $W = (m/l) dx \ g(x) \\$ So, total work done to put $l/3$ part back on the table $\\$ $W = \int_{0}^{\frac{1}{3}}(m/l)gx \ dx \Rightarrow w = (m/l)g \Bigg[ \frac{x^2}{2} \Bigg]_0 ^\frac{l}{3} = \frac{mgl^2}{18l} = \frac{mgl}{18} $

**39.** A uniform chain of length $L$ and mass $M$ overhangs a
horizontal table with its two third part on the table. The
friction coefficient between the table and the chain is $11.$
Find the work done by the friction during the period the
chain slips off the table.

Let, x length of chain is on the table at a particular instant . $\\$ So, work done by frictional force on a small element 'dx' $\\$ $dW_f=\mu Rx = \mu \Bigg( \frac{M}{L}dx \Bigg) gx \qquad$ [where $dx = \frac{M}{L}dx$] $\\$ Total work done by friction, $\\$ $W_f$ = $$\int_{2L/3}^{0} \mu \frac{M}{L}gx \ dx$$ $\\$ $\therefore W_f = \mu \frac{M}{L}g \Bigg[ \frac{x^2}{2} \Bigg]_{2L/3}^0 = \mu \frac{M}{L} \Bigg[ \frac{4L^2}{18} \Bigg] = 2\mu Mg \ L/9$

**40.** A block of mass $1$ kg is placed at the point $A$ of a rough
track shown in figure (8-E6). If slightly pushed towards
right, it stops at the point $B$ of the track. Calculate the
work done by the frictional force on the block during its
transit from $A$ to $B.$

m = 5kg, x = 10cm = 0.1m, v = 2m/sec, $\\$ h = ? $G = 10m/sec^2 \\$ So, $k = \frac{mg}{x}= \frac{50}{0.1} = 500 N/m \\$ Total energy just after the blow $E = \frac{1}{2} \ mv^2 + \frac{1}{2} \ kx^2 \qquad$ ...(1) $\\$ Total energy a a height $h = \frac{1}{2} \ k (h-x)^2 + mgh \qquad $ ...(2) $\\$ $\frac{1}{2} \ mv^2 + \frac{1}{2} \ kx^2 = \frac{1}{2} \ k (h-x)^2 + mgh$ $\\$ on solving we can get , $\\$ H = 0.2 m = 20 cm

**41.** A block of mass $5.0$ kg is suspended from the end of a
vertical spring which is stretched by $10$ cm under the
load of the block. The block is given a sharp impulse
from below so that it acquires an upward speed of $2.0$
$m/s.$ How high will it rise ? Take $g$ = $10$ $m/s^2$

**42.** A block of mass $250$ g is kept on a vertical spring of
spring constant $100$ $N/m$ fixed from below. The spring
is now compressed to have a length $10$ cm shorter than
its natural length and the system is released from this
position. How high does the block rise ? Take
$g$ =$10$ $m/s^2.$

m = 250 g = 0.250 kg, $\\$ k = 100 N/m, m = 10 cm = 0.1 m $\\$ $g = 10 m/sec^2 \\$ Applying law of conservation of energy $\\$ $\frac{1}{2} \ kx^2 = mgh \Rightarrow h = \frac{1}{2} \Bigg( \frac{ kx^2 }{mg} \Bigg) = \frac{100 \times (0.1)^2}{2 \times 0.25 \times 10} = 0.2m = 20 cm$

**43.** Figure (8-E7) shows a spring fixed at the bottom end of
an incline of inclination $37°.$ A small block of mass $2$ kg
starts slipping down the incline from a point $4.8$ $m$ away
from the spring. The block compresses the spring by
$20$ cm, stops momentarily and then rebounds through a
distance of $1$ m up the incline. Find (a) the friction
coefficient between the plane and the block and (b) the
spring constant of the spring. Take $g$ = $10$ $m/s^2.$

m = 2kg, $s_1 = 4.8m,$ x = 20cm = 0.2 m, $s_2 = 1m,$ $\\$ sin $37° = 0.60 = 3/5,$ $\theta = 37°,$ cos $37° = .79 = 0.8 = 4/5$ $\\$ $g = 10m/sec^2 \\$ Applying work-Energy principle for downward motion of the body$\\$ $0-0=mg \ sin \ 37°\times 5 - \mu R \times 5 - \frac{1}{2} \ kx^2$ $\\$ $\Rightarrow 20 \times (0.60) \times 1 - \mu \times 20 \times (0.80) \times 1 + \frac{1}{2} \ k \ (0.2)^2=0 \\$ $\Rightarrow 60-80 \mu - 0.02k = 0 \Rightarrow 80 \mu +0.02k=60 \qquad$ ...(1) $\\$ Similarly, for the upward motion of the body the equation is $\\$ $0-0= (-mg \ sin \ 37°) \times 1 - \mu R \times 1 + \frac{1}{2} \ k \ (0.2)^2 \\$ $\Rightarrow -20 \times (0.60) \times 1 - \mu \times 20 \times (0.80) \times 1 + \frac{1}{2} \ k \ (0.2)^2 = 0$ $\\$ $\Rightarrow -12-16 \mu + 0.02 K = 0 \qquad$ ...(2) $\\$ Adding equation (1) $\&$ equation (2), We get $96\mu = 48$ $\\$ $\Rightarrow \mu = 0.5$ $\\$ Now putting the value of $\mu$ in equation (1) K = 1000 N/m

**44.** A block of mass $m$ moving at a speed $v$ compresses a
spring through a distance $x$ before its speed is halved.
Find the spring constant of the spring.

44 None

SolutionsLet the velocity of the body at A be v $\\$ So,the velocity of the body at B is v/2 $\\$ Energy at point A = Energy at point B $\\$ So, $\frac{1}{2} \ mv_A^2 = \frac{1}{2} \ mv_B^2 + \frac{1}{2} \ kx^{2+}$ $\\$ $\Rightarrow \frac{1}{2} \ kx^2 = \frac{1}{2} \ mv_A^2 - \frac{1}{2} \ mv_B^2 \Rightarrow kx^2 = m (V_A^{2+-} V_B^2) \\ \Rightarrow kx^2 = m \Bigg(v^2 - \frac{v^2}{4} \Bigg)\Rightarrow k = \frac{3mv^2}{3x^2}$

**45.** Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.

45 None

Solutions**46.** Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.

46 None

Solutions**47.** Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.

47 None

Solutions**48.** Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.

48 None

Solutions**49.** Consider the situation shown in figure (8-E8). Initially
the spring is unstretched when the system is released
from rest. Assuming no friction in the pulley, find the
maximum elongation of the spring.

Mass of the body = m $\\$ Let the elongation be x $\\$ So, $\frac{1}{2} \ kx^2 = mgx \\$ $\Rightarrow x = 2mg/k$

**50.** A block of mass $m.$ is attached to two unstretched springs
of spring constants $k_1$ and $k_2$ as shown in figure (8-E9).
The block is displaced towards right through a distance $x$ and is released. Find the speed of the block as it passes
through the mean position shown.

The body is displaced x towards right $\\$ Let the velocity of the body be v at its mean position $\\$ Applying law of conservation of energy $\\$ $\frac{1}{2} \ mv^2 = \frac{1}{2} \ k_1 x^2 \Rightarrow mv^2 = x^2 (k_1+k_2) \Rightarrow v^2 = \frac{x^2(k_1+k_2)}{m} \\$ $\Rightarrow v = x \sqrt{\frac{k_1+k_2}{m}}$

**51.** A block of mass $m,$ sliding on a smooth horizontal surface
with a velocity $\vec v$ meets a long horizontal spring fixed at
one end and having spring constant $k$ as shown in figure
(8-E10). Find the maximum compression of tin spring.
Will the velocity of the block be the same as $\vec v$ when it
comes back to the original position shown ?

Let the compression be x $\\$ According to the law of conservation of the energy $\\$ $\frac{1}{2} \ mv^2 = \frac{1}{2} \ kx^2 \Rightarrow x^2 = mv^2/k \Rightarrow x= v \sqrt{(m/k)} \\$ b) No. it will be in the opposite direction and magnitude will be less due to loss in spring.

**52.** A small block of mass $100$ $g$ is pressed against a
horizontal spring fixed at one end to compress the spring
through $5.0$ $cm$ (figure 8-E11). The spring constant is
$100$ $N/m.$ When released, the block moves horizontally
till it leaves the spring. Where will it hit the ground $2$ $m$
below the spring ?

m = 100g = 0.1 kg, $\quad$ x = 5cm = 0.05m, $\quad$ k = 100 N/m $\\$ when the body leaves the spring, let the velocity be v $\\$ $\frac{1}{2} \ mv^2 = \frac{1}{2} \ kx^2 \Rightarrow v = x \sqrt{(k/m)} = 0.05 \times \sqrt{\frac{100}{0.1}} = 1.58 m/sec \\$ For the projectile motion, $\theta = 0^{\circ},$ Y = -2 $\\$ Now, y = (u sin $\theta$)t - $\frac{1}{2} \ gt^2 \\$ $ \Rightarrow -2 = (-1/2) \times 9.8 \times t^2 \Rightarrow t = 0.63 sec. \\$ So, x = (u cos $\theta$)t $\Rightarrow 1.58 \times 0.63 = 1m$

**53.** A small heavy block is attached to the lower end of a
light rod of length $l$ which can be rotated about its
clamped upper end. What minimum horizontal velocity
should the block be given so that it moves in a complete
vertical circle ?

Let the velocity of the body at A is 'V' for minimum velocity given at A velocity of the body at point B is zero. $\\$ Applying law of conservation of energy at A $\&$ B $\\$ $\frac{1}{2} \ mv^2 = mg \ (2l) \Rightarrow v = \sqrt{(4gl)} = 2 \sqrt{gl}$

**54.** Figure (8-E12) shows two blocks $A$ and $B,$ each having
a mass of $320$ $g$ connected by a light string passing over
a smooth light pulley. The horizontal surface on which
the block A can slide is smooth. The block $A$ is attached to a spring of spring constant $40$ $N/m$ whose other end
is fixed to a support $40$ $cm$ above the horizontal surface.
Initially, the spring is vertical and unstretched when the
system is released to move. Find the velocity of the block
$A$ at the instant it breaks off the surface below it. Take
$g$ = $10$ $m/s^2$

m = 320g = 0.32kg $\\$ k = 40N/m $\\$ h = 40cm = 0.4m $\\$ g = 10 m/s$^2 \\$ From the free body diagram, $\\$ $kx \ cos \theta = mg \\$ (when the block breaks off R = 0) $\\$ $\Rightarrow cos \theta = mg/kx \\$ So, $\frac{0.4}{0.4+x} = \frac{3.2}{40 \times x} \Rightarrow 16x = 3.2x +1.28 \Rightarrow x = 0.1 m \\$ So, $s = AB = \sqrt{(h+x)^2-h^2} = \sqrt{(0.5)^2-(0.4)^2} = 0.3m \\$ Let the velocity of the body at B be v $\\$ Charge in K.E. = work done (for the system) $\\$ $\frac{1}{2} \ mv^2 + \frac{1}{2} \ mv^2 = -\frac{1}{2} \ kx^2+mgs $ $\\$ $\Rightarrow (0.32) \times v^2 = - \frac{1}{2} \times 40 \times (0.1)^2 + 0.32 \times 10 \times (0.3) \\ \Rightarrow v= 1.5 m/s. $

**55.** One end of a spring of natural length $h$ and spring
constant $k$ is fixed at the ground and the other is fitted
with a smooth ring of mass $m$ which is allowed to slide
on a horizontal rod fixed at a height $h$ (figure 8-E13).
Initially, the spring makes an angle of $37°$ with the
vertical when the system is released from rest. Find the
speed of the ring when the spring becomes vertical.

$\theta = 37^{\circ}$ ; l = h = natural length $\\$ Let the velocity when the spring is vertical be 'v'. $\\$ Cos $37^{\circ}$ = BC/AC = 0.8 = 4/5 $\\$ AC = (h + x) = 5h/4 (because BC = h) $\\$ So, x = (5h/4) - h = h/4 $\\$ Applying work energy principle $\frac{1}{2} \ kx^2 = \frac{1}{2} \ mv^2 \\$ $\Rightarrow v= x \sqrt{(k/m)}= \frac{h}{4} \sqrt{\frac{k}{m}}$

**56.** Figure (8-E14) shows a light rod of length $l$ rigidly
attached to a small heavy block at one end and a hook
at the other end. The system is released from rest with
the rod in a horizontal position. There is a fixed smooth
ring at a depth $h$ below the initial position of the hook
and the hook gets into the ring as it reaches there. What
should be the minimum value of $h$ so that the block
moves in a complete circle about the ring ?

The minimum velocity required to cross the height point c = $\sqrt{2gl} \\$ Let the rod release from a height h. $\\$ Total energy at A = total energy at B $\\$ $mgh = \frac{1}{2} \ mv^2 ; mgh = \frac{1}{2} \ m \ (2gl) \\$ [ Because v = required velocity at B such that the block makes a complete circle.] $\\$ So, h = l.

**57.** The bob of a pendulum at rest is given a sharp hit to
impart a horizontal velocity. $\sqrt{10\ gl}$, where $1$ is the length
of the pendulum. Find the tension in the string when
(a) the string is horizontal, (b) the bob is at its highest
point and (c) the string makes an angle of $60°$ with the
upward vertical.

a) Let the velocity at B be $v_2 $ $\\$ $\frac{1}{2} \ mv_1^2 = \frac{1}{2} \ mv_2^2 + mgl $ $\\$ $\Rightarrow 1/2 \ m \ (10 \ gl)= \frac{1}{2} \ mv_2^2 + mgl$ $\\$ $v_2^2 = 8 \ gl $ $\\$ So, the tension in the string at horizontal position $\\$ $T = \frac{mv^2}{R}= \frac{m8gl}{l} = 8\ mg $ $\\$ b) Let the velocity at C be $V_3 $ $\\$ $\frac{1}{2} \ mv_1^2= \frac{1}{2} \ mv_3^2 + mg \ (2l) $ $\\$ $\Rightarrow \frac{1}{2} \ m \ (log l) = \frac{1}{2} \ mv_3^2 + 2mgl $ $\\$ $\Rightarrow v_3^2 = 6 \ mgl $ $\\$ So, the tension in the string is given by $\\$ $T_c = \frac{mv^2}{l} - mg = \frac{6glm}{l}mg = 5 \ mg $ $\\$ c) Let the velocity at point D be $v_4 $ $\\$ Again, $\frac{1}{2} \ mv_1^2 = \frac{1}{2} \ mv_4^2 + mgh $ $\\$ $\frac{1}{2} \times m \times (10 / gl) = 1.2 \ mv_4^2 + mgl (1 + cos 60^{\circ}) $ $\\$ $\Rightarrow v_4^2 = 7 \ gl $ $\\$ So, the tension in the string is $\\$ $T_D = (mv^2/l)-mg \ cos 60^{\circ} $ $\\$ $=m(7 \ gl)/l - l-0.5\ mg \Rightarrow 7 \ mg-0.5 \ mg = 6.5 \ mg.$

**58.** A simple pendulum consists of a $50$ $cm$ long string
connected to a $100$ $g$ ball. The ball is pulled aside so
that the string makes an angle of $37°$ with the vertical
and is then released. Find the tension in the string when
the bob is at its lowest position.

From the figure, $cos \theta = AC/AB $ $\\$ $\Rightarrow AC = AB \ cos \theta \Rightarrow (0.5) \times (0.8) = 0.4. $ $\\$ So, CD = (0.5) - (0.4) = (0.1) m $\\$ Energy at D = energy at B $\\$ $\frac{1}{2} \ mv^2 = mg (CD) $ $\\$ $v^2 = 2 \times 10 \times (0.1) = 2 $ $\\$ So, the tension is given by, $\\$ $T = \frac{mv^2}{r}+ mg = (0.1) \Big(\frac{2}{0.5}+10 \Big) = 1.4 \ N. $ $\\$

**59.** Figure (8-E15) shows a smooth track, a part of which is
a circle of radius $R.$ A block of mass m is pushed against
a spring of spring constant $k$ fixed at the left end and is then released. Find the initial compression of the
spring so that the block presses the track with a force
$mg$ when it reaches the point $P,$ where the radius of the
track is horizontal.

Given, N = mg $\\$ As shown in the figure, $mv^2/R=mg $ $\\$ $\Rightarrow v^2 = gR \qquad$ ...(1) $\\$ Total energy at point A = energy at P $\\$ $\frac{1}{2} \ kx^2 = \frac{mgR+2mgR}{2} \quad [because \ v^2 = gR] $ $\\$ $\Rightarrow x^2 = 3mgR/k \Rightarrow x = \sqrt{(3mgR)/k}$

**60.** The bob of a stationary pendulum is given a sharp hit
to impart it a horizontal speed of $\sqrt{3 \ gl}$
. Find the angle
rotated by the string before it becomes slack.

$V = \sqrt{3gl} $ $\\$ $\frac{1}{2} \ mv^2 - \frac{1}{2} \ mu^2 = -mgh $ $\\$ $v^2 = u^2 - 2gl(l+lcos \theta) $ $\\$ $\Rightarrow v^2 = 3gl -2gl (1+cos \theta) \qquad $ ...(1) $\\$ Again, $\\$ $mv^2/l = mg \ cos \theta $ $\\$ $v^2 = lg \ cos \theta $ $\\$ From equation (1) and (2) we get $\\$ $3gl - 2gl-2gl / cos \theta = gl \ cos \theta $ $\\$ $3 \ cos \theta = 1 \Rightarrow cos \theta = 1/3 $ $\\$ $\theta = cos^{-1}(1/3) $ $\\$ So, angle rotated before the string becomes slack $\\$ $= 180^{\circ} - cos^{-1} (1/3) = cos^{-1}(-1/3)$

**61.** A heavy particle is suspended by a $1.5$ $m$ long string. It
is given a horizontal velocity of $\sqrt{57}$
$m/s.$ (a) Find the
angle made by the string with the upward vertical, when
it becomes slack. (b) Find the speed of the particle at
this instant. (c) Find the maximum height reached
by the particle over the point of suspension. Take
$g$ =$10$ $m/s^2$

l = 1.5 m; $u = \sqrt{57} m/sec.$ $\\$ a) $mg \ cos \theta = mv^2 / l $ $v^2 = lg \ cos \theta \qquad$ ...(1) $\\$ change in K. E. = work done $\\$ $\frac{1}{2} \ mv^2 - \frac{1}{2} \ mu^2 = mgh $ $\\$ $\Rightarrow v^2 - 57 = -2 \times 1.5 g (1+cos \theta) \qquad$ ...(2) $\\$ $\Rightarrow v^2 = 57 - 3g (1+cos \theta) $ $\\$ Putting the value of v from equation (1) $\\$ $15 \ cos \theta = 57 - 3g(1+cos \theta) \Rightarrow 15 \ cos \theta = 57 - 30-30 \ cos \theta $ $\\$ $\Rightarrow 45 \ cos \theta = 27 \Rightarrow cos \theta=3/5. $ $\\$ $\Rightarrow \theta = cos^{-1} (3/5)= 53^{\circ} $ $\\$ b) $v = \sqrt{57- 3g (1+cos \theta)} \quad$ from equation (2) $\\$ $ = \sqrt{9} = 3 \ m/sec. $ $\\$ c)As the string becomes slack at point B, the particle will start making projectile motion. $\\$ $H= OE + DC = 1.5 \ cos \theta+ \frac{u^2sin^2 \theta}{2g} $ $\\$ $=(1.5 \times (3/5) + \frac{9 \times (0.8)^2}{2 \times 10} = 1.2 \ m.$

**62.** A simple pendulum of length $L$ having a bob of mass $m$
is deflected from its rest position by an angle $\theta$ and
released (figure 8-E16). The string hits a peg which is
fixed at a distance $x$ below the point of suspension and
the bob starts going in a circle centred at the peg. (a)
Assuming that initially the bob has a height less than
the peg, show that the maximum height reached by the
bob equals its initial height. (b) If the pendulum is
released with $\theta = 90°$ and $x=L/2$ find the maximum
height reached by the bob above its lowest position
before the string becomes slack. (c) Find the minimum
value of $x/L$ for which the bob goes in a complete circle
about the peg when the pendulum is released from
$\theta = 90°.$

a) When the bob has an initial height less that the peg and then released from rest (figure 1), $\\$ let body travels from A to B. $\\$ Since, Total energy at A = Total energy at B $\\$ $ \therefore (K.E.)_A = (PE)_A = (KE)_B + (PE)_B $ $\\$ $(PE)_A = (PE)_B \quad [because , (KE)_A = (KE)_B = 0] $ $\\$ So, the maximum height reached by the bob is equal to initial height. $\\$ b) When the pendulum is released with $\theta = 90^{\circ} \ and \ x = \frac{L}{2}$ (figure 2) the path of the particle $\\$ is shown in the figure 2. $\\$ At point C, the string will become slack and so the particle will start making projectile motion. $\\$ $\frac{1}{2} \ mv_c^2-0= mg \ (L/2) \ (1- cos \ \alpha) $ $\\$ because, distance between A and C in the verticle direction is $L/2 (1- cos \ \alpha) $ $\\$ $\Rightarrow v_c^2 = gL(1-cos \ \theta) \qquad$ ...(1) $\\$ Again, from the free body diagram $\\$ $\frac{mv_c^2}{L/2} = mg \ cos \ \alpha \ \big\{ because T_c = 0 \big\} $ $\\$ So, $V_c^2 = \frac{gL}{2} cos \ \alpha \qquad$ ...(2) $\\$ From Eqn. (1) and equn (2), $\\$ $gL(1- cos \ \alpha) = \frac{gL}{2} \ cos \ \alpha $ $\\$

$\Rightarrow 1-cos \ \alpha = \frac{1}{2} \ cos \ \alpha $ $\\$ $\Rightarrow \frac{3}{2} \ cos \ \alpha=1 \Rightarrow cos \ \alpha = 2/3 \quad$ ...(3) $\\$ To find highest position C, before the string becomes slack $\\$ $BF = \frac{L}{2} + \frac{L}{2}cos \ \theta = \frac{L}{2} + \frac{L}{2} \times \frac{2}{3} = L \Big(\frac{1}{2}+ \frac{1}{3} \Big) $ $\\$ So, BF = (5L/6) $\\$ c) If the particle has to comlpete a vertical circle, at the point C. $\\$ $\frac{mv_c^2}{(L-x)} = mg $ $\\$ $\Rightarrow V_c^2 = g \ (L-x) \qquad$ ...(1) $\\$ Again applying energy principle between A and C, $\\$ $\frac{1}{2} \ mv_c^2 - 0 = mg \ (OC) $ $\\$ $\Rightarrow \frac{1}{2} \ v_c^2 = mg [L-2(L-x)] = mg \ (2x -L) $ $\\$ $\Rightarrow v_c^2 = 2g(2x - L) \qquad$ ...(2) $\\$ From equn (1) and equn (2) $\\$ $g(L-x) = 2g \ (2x - L) $ $\\$ $\Rightarrow L-x = 4x-2L $ $\\$ $\Rightarrow 5x = 3L $ $\\$ $\therefore \frac{x}{L}= \frac{3}{5}=0.6 $ $\\$ So, the rates (x/L) should be 0.6

**63.** A particle slides on the surface of a fixed smooth sphere
starting from the topmost point. Find the angle rotated
by the radius through the particle, when it leaves contact
with the sphere.

Let the velocity be v when the body leaves the surface. $\\$ From the free body diagram $\\$ $\frac{mv^2}{R} = mg \ cos \ \theta$ [Because normal reaction] $\\$ $v^2 = Rg \ cos \ \theta \qquad$ ...(1) $\\$ Again, from work-energy principle, $\\$ Change in K.E. = work done $\\$ $\Rightarrow \frac{1}{2} \ mv^2 - 0=mg(R-R \ cos \ \theta) $ $\\$ $v^2 = 2gR \ (1-cos \ \theta) \qquad$ ...(2) $\\$ From (1) and (2) $\\$ $Rg \ cos \ \theta = 2gR(1- cos \ \theta) $ $\\$ $3gR \ cos \ \theta = 2 \ gR $ $\\$ $cos \ \theta = 2/3 $ $\\$ $\theta = cos^{-1} (2/3) $ $\\$

**64.** A particle of mass $m$ is kept on a fixed, smooth sphere
of radius $R$ at a position, where the radius through the
particle makes an angle of $30°$ with the vertical. The
particle is released from this position. (a) What is the
force exerted by the sphere on the particle just after the
release ? (b) Find the distance travelled by the particle
before it leaves contact with the sphere.

a) When the particle is released from rest, the centrifugal force is zero. $\\$ N force is zero = $mg \ cos \ \theta $ $\\$ $= mg \ cos \ 30^{\circ} = \frac{\sqrt{3}mg}{2} $ $\\$ b) When the prticle is leaves contact with the surface , N = 0. $\\$ So, $\frac{mv^2}{R} mg \ cos \ \theta $ $\\$ $\Rightarrow v^2 = Rg \ cos \ \theta \qquad$ ...(1) $\\$ Again, $\frac{1}{2} \ mv^2 = mgR (cos \ 30^{\circ} - cos \ \theta) $ $\\$ $\Rightarrow v^2 = 2Rg \Bigg( \frac{\sqrt{3}}{2}-cos \ \theta \Bigg) \qquad$ ...(2) $\\$ From equn (1) and equn (2) $\\$ $Rg cos \ \theta = \sqrt{3} \ Rg-2Rg \ cos \ \theta $ $\\$ $\Rightarrow 3 \ cos \ \theta = \sqrt{3} $ $\\$ $\Rightarrow cos \ \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = cos^{-1} \Bigg( \frac{1}{\sqrt{3}} \Bigg) $ $\\$ So, the distance travelled by the particle before leaving contact, $\\$ $l = R(\theta - \pi/6) \quad [Because \ 30^{\circ} = \pi/6] $ $\\$ putting the value of $\theta$ we get $l = 0.43R$

**65.** A particle of mass $m$ is kept on the top of a smooth
sphere of radius $R.$ It is given a sharp impulse which
imparts it a horizontal speed $v.$ (a) Find the normal force
between the sphere and the particle just after the
impulse. (b) What should be the minimum value of $v$ for
which the particle does not slip on the sphere ?
(c) Assuming the velocity $v$ to be half the minimum
calculated in part, (d) find the angle made by the radius through the particle with the vertical when it leaves the
sphere.

a) Radius = R $\\$ horizontal speed = v $\\$ from the free body diagram, $\\$ N = Normal force = $mg-\frac{mv^2}{R} $ $\\$ b) when the particle is given maximum velocity so that the centrifugal force balance the weight, the particle does not slip on the sphere. $\\$ $\frac{mv^2}{R} = mg \Rightarrow v = \sqrt{gR} $ $\\$ c) If the body is given velocity $v_1 $ $\\$ $v_1 = \sqrt{gR}/2 $ $\\$ $v_1^2 - gR/4 $ $\\$ Let the velocity be $v_2$ when it leaves contact with the surface,$\\$ So, $\frac{mv^2}{R} = mg \ cos \ \theta $ $\\$ $\Rightarrow v_2^2 = Rg \ cos \ \theta \qquad$ ...(1) $\\$ Again, $\frac{1}{2} \ mv_2^2 - \frac{1}{2} \ mv_1^2 = mgR \ (1- cos \ \theta )$ $\\$ $\Rightarrow v_2^2 = v_1^2+2gR \ (1- cos \ \theta) \qquad$ ...(2) $\\$ From equn (1) and equn (2) $\\$ $Rg \ cos \ \theta = (Rg/4) + 2gR \ (1-cos \ \theta ) $ $\\$ $\Rightarrow cos \ \theta = (1/4) + 2 - 2 \ cos \ \theta $ $\\$ $\Rightarrow 3 \ cos \ \theta = 9/4 $ $\\$ $\Rightarrow cos \ \theta = 3/4 $ $\\$ $\Rightarrow \theta = cos^{-1} (3/4)$

**66.** Figure (8-E17) shows a smooth track which consists of
a straight inclined part of length $1$ joining smoothly with
the circular part. A particle of mass $m$ is projected up
the incline from its bottom. (a) Find the minimum
projection-speed $v_o$ for which the particle reaches the top
of the track. (b) Assuming that the projection-speed is
$2v_o$ and that the block does not lose contact with the
track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the
projection-speed is only slightly greater than $v_o$, where
will the block lose contact with the track ?

a) Net force on the particle between A and B, F = $mg \ sin \ \theta $ $\\$ work done to reach B, $W = FS=mg \ sin \ \theta \ l $ $\\$ Again work done to reach B to C $= mgh = mg \ R \ (1-cos \ \theta) $ $\\$ So, total work done $= mg[l sin \ \theta + R(1-cos \ \theta)] $ $\\$ Now, change in K.E. = work done $\\$ $\Rightarrow \frac{1}{2} \ mv_o^2=mg \ [l \ sin \ \theta +R \ (1-cos \theta)] $ $\\$ $\Rightarrow v_o = \sqrt{2g(R(1- cos \ \theta)+l \ sin \ \theta)} $ $\\$ b) When the block is projected at a speed $2v_o. $ $\\$ Let the velocity at C will be $V_c. $ $\\$ Applying energy principle, $\\$ $\frac{1}{2} \ mv_c^2-\frac{1}{2} \ m \ (2v_o)^2 = -mg \ [l \ sin \ \theta+ R(1 - cos \ \theta)] $ $\\$ $= v_c^2 = 4v_o-2g \ [l \ sin \ \theta + R(1-cos \ \theta)]$ $\\$ $4.2g \ [l \ sin \ \theta + R(1-cos \ \theta)]-2g \ [l \ sin \ \theta + R(1- cos \ \theta)] $ $\\$ $ = 6g \ [l sin \ \theta + R(1 - cos \ \theta)] $ $\\$ So, force acting on the body, $\\$ $\Rightarrow N = \frac{mv_c^2}{R} = 6 mg \ [(l/R) \ sin \ \theta + 1-cos \ \theta] $ $\\$ c) Let the block loose contact after making an angle $\theta $ $\\$ $\frac{mv^2}{R} = mg \ cos \ \theta \Rightarrow v^2 = Rg \ cos \ \theta \qquad$ ...(1)$\\$ Again, $\frac{1}{2} \ mv^2 = mg \ (R - R \ cos \theta) \Rightarrow v^2 = 2gR \ (1-cos \ \theta) \qquad$ ...(2) $\\$ From (1) and (2) cos $\theta$ = 2/3 $\Rightarrow \theta = cos^{-1}(2/3)$

**67.** A chain of length $l$ and mass $m$ lies on the surface of a
smooth sphere of radius $R >l$ with one end tied to the
top of the sphere. (a) Find the gravitational potential
energy of the chain with reference level at the centre of
the sphere. (b) Suppose the chain is released and slides
down the sphere. Find the kinetic energy of the chain,
when it has slid through an angle $\theta.$ (c) Find the
tangential acceleration $\frac{dv}{dt}$ of the chain when the chain
starts sliding down.

Let us consider a small element which makes angle '$d\theta$' at the centre. $\\$ $\therefore dm = (m/l)Rd \ \theta$ $\\$ a) Gravitational potential energy of 'dm' with respect to centre of the sphere $\\$ $= (dm)g \ R \ cos \ \theta$ $\\$ $(mg/l) \ R \ cos \ \theta \ d\theta$ $\\$ So, Total, G.P.E. $$= \int_{0}^{l/r} \frac{mgR^2}{l} cos \ \theta \ d\theta \qquad [ \ \alpha = (l/R)]$$(angle subtended by the chain at the centre)......$\\$ $=\frac{mR^2g}{l}[sin \ \theta](l/R)=\frac{mRg}{l}sin \ (l/R)$ $\\$ b) When the chain is released from rest and slides down through an angle $\theta$, the K.E. of the chain is given $\\$ K.E. = Change in potential energy. $\\$ $=\frac{mR^2g}{l} sin (l/R)-m $ $$\int \frac{gR^2}{l} cos \ \theta \ d \ \theta......?$$ $\\$ $= \frac{mR^2g}{l}\ \bigg[ sin (l/R) + sin \ \theta - sin \ \big\{\theta + (l/R) \big\} \bigg]$ $\\$ c) Since, $K.E. = \frac{1}{2} \ mv^2=\frac{mR^2g}{l} \ \bigg[sin (l/R) + sin \ \theta - sin \ \big\{\theta + (l/R) \big\} \bigg] $ $\\$ Taking derivative of both sides with respect to 't' $\\$ $(1/2) \times 2v \times \frac{dv}{dt} = \frac{R^2g}{l} \ \Big[ cos \ \theta \times \frac{d\theta}{dt}-cos(\theta+l/R) \frac{d\theta}{dt} \Big]$ $\\$ $\therefore (R \ \frac{d\theta}{dt}) \ \frac{dv}{dt}= \frac{R^2g}{l} \times \frac{d\theta}{dt} \ \Big[ cos \ \theta -cos(\theta+(l/R)) \Big] $ $\\$ When the chain start sliding down, $\theta=0.$ $\\$ So, $\frac{dv}{dt} = \frac{Rg}{l} \ [1-cos \ (l/R)]$ $\\$

**68.** A smooth sphere of radius $R$ is made to translate in a
straight line with a constant acceleration $a.$ A particle
kept on the top of the sphere is released from there at
zero velocity with respect to the sphere. Find the speed
of the particle with respect to the sphere as a function
of the angle $\theta$ it slides

Let the sphere move towards left with an acceleration 'a' $\\$ Let m = mass of the particle $\\$ The particle 'm' will also experience the inertia due to acceleration 'a' as it is on the sphere. It will also experience the tangential inertia force (m (dv/dt)) and centrifugal force $(mv^2/R).$ $\\$ $m \frac{dv}{dt} = ma \ cos \ \theta + mg \ sin \ \theta \Rightarrow mv \ \frac{dv}{dt} \\ = ma \ cos \ \theta \ \Big( R \frac{d\theta}{dt} \Big) + mg \ sin \ \theta$ $\\$ $\Big( R \frac{d\theta}{dt} \Big)$ $\\$ Because, $v = R \frac{d\theta}{dt}$ $\\$ $\Rightarrow vd \ v = a \ R \ cos \ \theta \ d\theta + gR \ sin \ \theta \ d\theta$ $\\$ Integrating both sides we get, $\\$ $\frac{v^2}{2}= a \ R \ sin \ \theta - gR \ cos \ \theta + C$ $\\$ Given that, at $\theta=0, v= 0, \ So, \ C = gR$ $\\$ So, $\frac{v^2}{2} = a \ R \ sin \ \theta - gR \ cos \ \theta + gR$ $\\$ $\therefore v^2 = 2R \ (a \ sin \ \theta + g-g \ cos \ \theta )$ $\\$ $\Rightarrow v=[2R \ (a \ \sin \theta) + g-g \ cos \ \theta)]^ {1/2} $