**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. $