1. WPE

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 1

EXERCISE # 11.2 Let tension in string be T, then work done by tension T = Td

Applying newtons second law on the bucket

ekuk jLlh esa ruko T gS] rks ruko T }kjk fd;k x;k dk;Z = Td

ckYVh ij U;wVu dk f}rh; fu;e yxkus ij]Mg T = M

a

gor T =

4

3Mg

required work done = 4

3Mg d

fd;k x;k vko';d dk;Z = 4

3Mg d

1.4 Change in velocity =mass

graphTFunderarea

=5

)10(40 = 6 m/s

WF

= K.E. =2

1(5) 62 = 90 J

1.5 BA

F/////////////////////

Consider the blocks shown in the figure to be moving together due to friction between them.

The free body diagrams of both the blocks is shown below.

fp=kkuqlkj ekuk nks CykWd ?k"kZ.k ds dkj.k lkFk&lkFk xfr dj jgs gSAnksauks CykWd ds FBD fp=k esa n'kkZ;s x;s gSA

Work done by static friction on A is positive and on B is negative.

A ij LFkSfrd ?k"kZ.k }kjk fd;k x;k dk;Z /kukRed rFkk B ij _.kkRed gSA

2.3 (D)W = F.ds= dssk

s

s1

= kln (s/s1)

3.3 (C)mg

2

11 =

2

1mv2

2

g=

2

v2 v = g

4.2 P =dt

d(mgh)

Pact

=5.0

100101000 P

act= 2000 kW P

consumption=

25.0

2000kW = 8000 kW.

4.3 At any time fdlh {k.k

v

= [(u cos) i + (u sin gt)j ]

P = j)gtsinu(icosu).jmg(v.F

= mg2 t mgu sin

Hence, power varies linearly with t ime.

vr% 'kfDr le; ds lkFk js[kh; :i ls ifjofrZr gksxhA

WORK POWER ENERGY

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4.7 (D)Instantaneous power delivered = P = v.F

= Fv

where, F f = ma

F = f + ma P = (f + ma) v

Put f = mg P = (mg + ma)v = m(a + g).at

F

f

am

v

4.8 The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below.

Hence as t increases, this area increases.

Work done by force keeps on increasing.

5.1* Since ;W = dr.F

Clearly for forces (A) and (B) the integration do not require any information of the path taken.

For (C) : Wc

=

)jdyidx(.)yx(

)jyix(32/322

=

2/322 )yx(

dyydxx3

Taking : x2 + y2 = t2xdx + 2y dy = dt

xdx + ydy =2

dt W

c= 2/32/3 t

dt

2

3

t

2/dt3

which is solvable.

Hence (A), (B) and (C) are conservative forces.

But (D) requires some more information on path. Hence non-conservative.

pwafd ; W = dr.F

(A) rFkk (B)

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5.2* Only the following statements are true from definition of a conservative force.

"Its work is zero when the particle mov es exactly once around any closed path".

"Its work depends on the end points of the motion, not on the path between".

fuEu dFku laj{kh cy dh ifjHkk"kk ls lR; gS A ;g gSA"tc d.k fdlh can iFk ds vuqfn'k Bhd ,d pDdj iwjk djrk gS rks laj{kh cy }kjk fd;k x;k dk;Z 'kwU; gksrk gSA "

" blds fy, fd;k x;k dk;Z xfr ds vfUre fcUnqvksa ij fuHkZj djrk gS iFk ij ughaA "

5.3 The force is constant and hence conservative

cy fu;r gS rFkk blfy;s laj{kh; (conservative) gS] W

1= W

2

6.2 x = x1

and x = x3

are not equilibrium positions becausedx

du 0 at these points.

x = x2

is unstable, as U is maximum at this point.

x = x1rFkk x = x

3lkE;koLFkk esa ugha gS D;ksafd bu fcUnqvksa ij

dx

du 0

x = x2vLFkkbZ lkE;koLFkk gS D;ksafd bl fcUnq ij U vf/kdre gSA

6.3 The work done by man is negative of magnitude of decrease in potential energy of chain

bl f;k esa O;fDr }kjk fd;k x;k dk;Z xq:Ro }kjk fd;s x;s dk;Z ds _.kkRed ds cjkcj gksrk gSA

L/4

L/2

U = mg2

L

2

mg

4

L= 3 mg

8

L W =

8

mg3

6.5 The two springs have different spring constants. Also energy remains conserved during the motion as no

friction is prestent.

6.8 (B)Apply COE; mgx=2

1kx2 x = 2mg/k

6.9* U = 3x + 4y

ay = m

Fy=

m

x/U = 3

ax = m

Fy

=

m

y/U

= 4 a

= 5 m/s2

Let at time 't' particle crosses y-axis

then 6 =2

1( 3) t2 t = 2 sec.

Along y-direction :

y =2

1( 4) (2)2 = 8

particle crosses y-axis at y = 4At (6, 4) : U = 34 & KE = 0

At (0, 4) : U = 16 KE = 50

or,2

1mv2 = 50 v = 10 m/s while crossing y-axis

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6.10 From conservation of energy tkZ laj{k.k ls

K.E. + P.E. = E or ;k K.E. = E 2

1kx2

K.E. at x = kE2

is

E 2

1k

k

E2= 0

The speed of particle at x = k

E2is zero.

x = k

E2ij d.k dh pky 'kwU; gSA

EXERCISE # 2PART - I

7. When the cart maximally compresses the spring net force on cart is up the incline and its speed is zero.

tc xkM+h] fLizax dks vf/kdre laihfM+r djsaxs rc xkM+h ij ifj.kkeh cy ry ds vuqfn'k ij dh rjQ gksxk rFkk pky 'kwU; gksxhA8. Case-I :Ground is smooth :

force on each block is same Kx0in the same direction, hence friction between them is zero. Hence,

work done by friction on block A, B or system is zero.

Case-II : Ground is rough :

f1

kx0A

f1

kx0Bf2

Obviously, workdone by friction on block A is negative. Since values of f1and f

2are unknown, hence workdone by

frictino on block B can't be determined.

9. Because the acceleration of wedge is zero, the normal reaction exerted by wedge on block is

N = mg cos37 .

The acceleration of the block is g sin 37 along the incline and initial velocity of the block is v = 10 m/s

horizontally towards right as shown in figure.

10. Minimum work done to accelerate the truck from speed 0 to v and from v to 2v are

Vd dks pky 0 ls v rd rFkk v ls 2v rd Rofjr djus ds fy, U;wure fd;s x;s dk;Z gSa &

W1

=21 mv2

21 m(0)2 =

21 mv2

and (rFkk) W2

=2

1m(2v)2

2

1m(v)2 =

2

3mv2

W1

< W2

The component of velocity of the block normal to the incline is v sin 37. Hence the displacement of the block

normal to the incline in t = 2 second is

S = v sin 37 2 = 10 5

3 2 = 12 m.

The work done by normal reaction

W = mg cos 37 S = 100 54 12 = 960 J

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11. Accelerat ion as shown in the graph can be converted into force by multi plying with m = 3 kg. Therefore

area under the curve (F x curve) is

iznf'kZr xzkQ ds Roj.k dks m = 3 kg ls xq.kk djds cy esa ifjofrZr dj ldrs gSA vr% oz (F x oz) ls ifjc) {ks=kQy

[2

1 2 12] + [4 12] = 60 J.

15. Let F the force with which man pulls the block.

Fv = 500 F = 50 N(F - mg) v = 100

solving m = 4 kg

16. As long as the block of mass m remains stationary, the block of mass M released from rest comes down by

K

Mg2(before coming it rest momentanly again).

Thus the maximum extension in spring is

x =KMg2 ................. (1)

for block of mass m to just move up the incline

kx = mg sin + mg cos ................. (2)

2Mg = mg 5

3+

4

3mg

5

4or M =

5

3m Ans.

18. System is block & string. Applying work energy theorem on system

CykWd rFkk jLlh fudk; gS] fudk; ij dk;Z tkZ izes; }kjk

(200)10 10g(R R cos60) = 2

1(10)v2

2(200 10 5) = v2

v = 300 = 310 .

20.

The above graphs show v t graph from a t graph & Then v2 t graph, which are self explanatory.

21. As ; Wext

= (ME) ; ME = Mechanical energy.Mechanical energy will keep on increasing upto the instant the W ext is positive, which will happen till there

is no compression in the spring. First the spring gets extended to a maximum and after which the exten-

sion decreases upto the natural length. After that there is a compression in the spring, results in a ve

external work (so as to move the end of spring at constant speed u).

Hence maximum energy stored is at the natural length.

& MEmax = 2

1mv 2

At the natural length v = 2u, since the block is moving at this instant at a speed u with respect to the other

end of the spring.

Hence MEmax = 21 m(2u)2 = 2mu2.

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pwafd Wext = (ME) ; ME = ;kf+U=kd tkZ;kfU=kd tkZ] Wext/kukRed gksus rd c

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vr% xfrt mtkZ vf/kdre gksxh tc 2M nzO;ekuK

Mg2foLFkkfir gksrk gSA

Applying W/E theorem (dk;Z tkZ izes; ls)

kf 0 = 2Mg. K

Mg2

2

22

K

gM4

.K2

1

kf = 2

22

K

gM2

Maximum energy of spring =

2

K

Mg4.K

2

1

=

K

gM8 22

fLizax dh vf/kdre tkZ =2

K

Mg4.K

2

1

=

K

gM8 22

Therefore Maximum spring energy = 4 maximum K.E.

vr% fLizax dh vf/kdre tkZ = 4 vf/kdre xfrt tkZ

When K.E. is maximum x =K

Mg2.

tc xfrt tkZ vf/kdre gS x =K

Mg2.

Spring energy (fLizax tkZ) =2

22

K

gM4.K.

2

1=

2

22

K

gM2

i.e. (D) is wrong.

vFkkZr~ (D) xyr gSA

PART - II

1.2

1mv2 =

2

1kx2

2

1m

m

k

2

3

2

=

2

1kx2

x =3

2=

cos

cos =5

3 = 53

2. Applying work-energy theorem between A and B.

A rFkk B e/; dk;Z mtkZ izes; ls

3m

3m

37

2

1mVB

2 2

1mVA

2 = Wgravity + W friction

2

1mVB

2 2

1mVA

2 = Wxq:Ro + W?k"kZ.k

2

1mVB

2 2

1m (136) = mg(3 + 3 sin 37) mg cos 37 x 3

2

V2B

2

136

= 48 12 VB = 4 m/s

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3. Area under Px graph = dxp = dxvdtdv

m

=

v

1

2 dVmv =

v

1

3

3

mv

=

37

10

(v3 1)

from graph ; area =

2

1(2 + 4) 10 = 30

Px xzkQ ds uhps dk {ks=kQy = dxp = dtdv

mvdx =

v

1

2 dVmv =

v

1

3

3

mv

=

37

10

(v3 1)

xzkQ ls ; {ks=kQy =2

1(2 + 4) 10 = 30

37

10

(v3 1) = 30

v = 4 m/s

ALITER : oSdfYid

from graph xzkQ lsP = 0.2 x + 2

or;k mvdx

dvv = 0.2 x + 2

or ;k mv 2 dv = (0.2 x + 2) dx

Now integrate both sides, nksuksa rjQ lekdyu djus ij v

1

2dvmv = 10

1

dx)2x2.0( v = 4 m/s.

4. As string does no work on the ball, energy conservation can be applied.

pwfd jLlh xsan ij dksbZ Hkh dk;Z ugh djrh gS vr% mtkZ laj{k.k ykxw dj ldrs gSA

2

1mV2 = mg (L L cos)

V = )cos1(Lg2

on putting values V = 10 m/s

eku j[kus ij V = 10 m/s

6. From work energy theorem d.k dh ij dh vksj xfr ds fy, dk;Z tkZ izes; yxkus ijfor upward motion

2

1m (16)2 = mgh + W (work by air resistance)

for downward motion uhps dh vksj xfr ds fy,

2

1m (8)2 = mgh W

2

1[(16)2 + (8)2] = 2 gh or h = 8 m

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EXERCISE # 31.1 The displacement of A shall be less than displacement L of block B.

Hence work done by friction on block A is positive and its magnitude is less than mgL.And the work done by friction on block B is negative and its magnitude is equal to mgL.Therefore workdone by friction on block A plus on block B is negative its magnitude is less than mgL.

Work done by F is positive. Since F>mg, magni tude of work done by F shall be more than mgL.Adk foLFkkiu CykWd Bds foLFkkiu Lls de gksxkA CykWdAij ?k"kZ.k }kjk fd;k x;k dk;Z /kukRed gS vkSj bldk ifjek.k mgLls de gSA CykWd B ij ?k"kZ.k }kjk fd;k x;k dk;Z _.kkRed gS vkSj bldk ifjek.k mgLds cjkcj gSA blfy;s CykWdA CykWdB ij ?k"kZ.k }kjk fd;k x;k dk;Z _.kkRed gSA bldk ifjek.k mgL ls de gSA

F }kjk fd;k dk;Z /kukRed gSA pwafd F > 2mg , F }kjk fd;k x;k dk;Z dk ifjek.k 2mgL ls vf/kd gksxk

EXERCISE # 4PART - I

1. Power 'kfDr P = F

. V

= FV

F = V

dt

dm

= V

dt

volume(d

dt

(d ruvk; = density ?kuRo

= V

dt

volume(d

dt

)(d ruvk;

= V (AV)= AV2

Power'kfDr P = AV3

or P V3

Alternate SolutionoSdfYid gy

Power output is proportional to number of molecular striking the blades per unit time [which depends onthe velocity V of wind] and also proportional to energy to striking molecules or proportional to square of

velocity V2 Therefore, power output P V3

fuxZr 'kfDr ,dkad le; esa CysM ls Vdjkus okys v.kqvksa dh la[;k ds lekuqikrh gksrh gS [tks ok;q ds osx V ij fuHkZjgS] rFkk Vdjkus okys v.kqvksa dh tkZ ds H kh lekuqikrh gksrh gS ;k osx ds oxZ ds lekuqikrh gksrh gS] vr% fuxZr 'kfDrP V3

2. F = dx

dU

dU = F . dx or ;k U(x) =

x

0

3 dx)axkx(

U(x) =2

kx2 4

ax4

U(x) = 0 and vkSj x = 0 and vkSj x =a

k2

U(x) = negative for _.kkRed x >a

k2

From the given function we can see that

F = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0. Therefore, the most appropriate option is (D).

fn;s x;s xzkQ ls ge ns[k ldrs gSa fd x = 0 ij F = 0 gS vFkkZr~ U-x xzkQ dh izo.krk x = 0 ij 'kwU; gSA vr% lokZf/kd mi;qZDrfodYi (D) gSA

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3. Let x be the maximum extension of the spring. From conservation of mechanical energy :

decrease in gravitational potential energy = increase in elastic potential energy

ekuk x fLizax esa vf/kdre f[kapko gSA ;kaf=kd tkZ laj{k.k ls :xq:Roh; fLFkfrt tkZ esa deh = izR;kLFk fLFkfrt tkZ es a o f)

Mgx = 21 kx2

or ;k x =k

Mg2

4. From F =dx

dU ls

x

0

x

0

)x(U

0

dx)kx(FdxdU

U(x) =2

kx2

as tSls U(0) = 0Therefore, the correct option is (A). vr% (A) lgh fodYi gSA

5. In horizontal plane Kinetic Energy of the block is completely converted into heat due to Friction but in the case

of inclined plane some part of this Kinetic Energy is also convert into gravitational Potential Energy. So decrease

in the mechanical energy in second situation is smaller than that in the first situation. So statement-1 is correct.

Cofficient of Friction does not depends on normal reaction, In case normal reaction changes with inclinationbut not cofficient of f riction so this statement is wrong.

Sol. {kSfrt ry esa xqVds dh xfrt tkZ iw.kZ:is.k ?k"kZ.k ds dkj.k mRiUu m"ek esa cny tkrh gS] ijUrq ur ij xfrt tkZ dk dqN

Hkkx xq:Roh; fLFkfrt mtkZ esa Hkh ifjofrZr gks tkrk gSA vr% ;kaf=kd tkZ esa deh f}rh; fLFkfr esa] izFke fLFkfr ls de gSA vr%dFku -1 lR; gS?k"kZ.k xq.kkad vfHkyEc izfrf;k cy fuHkZj ugha djrk gSA f}rh; fLFkfr esa vfHkyEc izfrf;k cy ur ry ds lkFk cny tkrkgS] ijUrq ?k"kZ.k xq.kkad ugha cnyrk gS] vr% ;s dFku vlR; gSA

6.

As springs and supports (m1

and m2) are having negligible mass. Whenever springs pull the massless

supports, springs will be in natural length. At maximum compression, velocity of B will be zero.

pwafd fLizax o lgk;dks (m1o m

2) ds ux.; nzO;eku gSA tc dHkh Hkh fLizax mUgs [khaprh gS rks izkdfrd yEckbZ esa gksxh]

vf/kdre laihMu ij B dk osx 'kwU; gksxk

And by energy conservation. vkSj tkZ laj{k.k ls

2

1

(4K) y2

= 2

1

Kx2

2

1

x

y

Ans. (C)

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7. T = gmm

mm2

21

21

=36.072.0

36.072.02

10

T = 4.8 N

a =g

mm

mm

21

21

= 3g

s =2at

2

1=

2

1

3

g

(1)2 =6

10

Work done by T = (T) (S)

T }kjk fd;k x;k dk;Z = (T) (S)

= (4.8) 6

10

= 8 J Ans.

8. pFdt

21 4 3

21 1.5 2 = pf 0 pf= 6 1.5 = 2

9

K.E. =m2

p2=

224

81

;K.E. = 5.06 J Ans.

9. By WET

0 1

2mv2 =

1

2

kx2 mgx

1

2 0.18 v2 =

1

2 2 (0.06)2 + 0.1 0.18 10 0.06

v = 0.4

N = 4

PART - II

1. Let initial velocity is u and retardation is a (ekuk izkjfEHkd osx u rFkk eanu a gS

So, (vr%)4

u2= u2 2a (0.03) ...(i)

0 = 4

u2

2a S ..(ii)

here S is required distance (;gk S vko';d nwjh gSfrom equation (i) & (ii) lehdj.k (i) o (ii) lsS = 0.01 m = 1 cm

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2. WC

= U= (U

final U

initial)

=

22 5k

2

115k

2

1]

WC = 8 Joule

3. K = 5 103 N/m

x = 5 cm

W1

=2

1k 21x = 2

15 103 (5 102)2

= 6.25 J

W2

=2

k(x

1+ x

2)2

= 2

5 103

(5 + 102

+ 5 102

)2

= 25J

Net work done (ifj.kkeh dk;Z) = W2

W1

= 25 6.25 = 18.75 J

= 18.75 N-m

4. v2 = u2 + 2ax

v2 = 2ax a =

x2

v2

= m.x2

v2.v =

x2

m 3v

v3 x (P = constant) (fu;r)v x1/3

dt

dxx1/3

dtdxx 3/1

2

3x2/3 t

xt3/2

5. Mass per unit length izfr bdkbZ yEckbZ dk nzO;eku

=L

M

=2

4= 2 kg/m

The mass of 0.6 m of chain 0.6 m pSu dk nzO;eku= 0.6 2 = 1.2 kg

The centre of mass of hanging part yVds gq, Hkkx dk nzO;eku dsUnz

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=2

06.0 = 0.3 m

Hence, work done in pulling the chain on the table vr% est ij psu dks [khapus esa fd;k x;k dk;ZW = mgh

= 1.2 10 0.3

= 1.2 10 0.3= 3. 6 J

6. Work done is displacing the particle d.k dks foLFkkfir djus esa fd;k x;k dk;Z

W = F

= r

= (5 i + 3j + 2 k ). (2 i j )

= 5 2 + 3 (1) + 2 0

= 10 3

= 7 J

7. Let the constant acceleration of body of mass m is a. ekuk m nzO;eku dh oLrq dk fu;r Roj.k a gSFrom equation of motion (xfr ds lehdj.k ls)

v1= 0 + at

1

a =1

1

t

vt

At an instant t, the velocity v of the body fdlh {k.k t ij d.k dk osx vv = 0 + at

v =1

1

t

vt

Therefore, instantaneous power

vr% rkR{kf.kd 'kfDrP = Fv= mav [F = ma]

= m

1

1

t

v

t.

t

v

1

1[from equations (i) and (ii)] [lehdj.k (i) o (ii) ls]

= 21

21

t

tmv

8. F = ma = T

m

T

0a

Instantaneous power (rkR{kf.kd 'kfDr) = F= ma

=T

m. at =

T

m.

T

. t

=2

2

T

m.t

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9. Maximum height attained by the particle

d.k }kjk izkIr vf/kdre pkbZ

m4

5

102

5

g2

uH

22

Wg = -MgH = -0.1 10 (5/4) = -1.25 J

10. Velocity of ball just after throwing

Qsadus ds Bhd i'pkr~ xsan dk osx

v = gh2 = 2102 = 40 m/s

Let a be the acceleration of ball during throwing, then

ekuk a Qsadus ds nkSjku xsan dk Roj.k gS] rks

v2 = u2 + 2as = 02 + 2as a =s2

v2=

2.02

40

= 100 m/s2

F - mg = ma F = m(g + a) = 0.2(10 + 100) = 22 N (2) is correct

11. kmv2

1 2 4K

mv2

1

4

1

4

vm

2

1)60cosv(m

2

1 22

2

12. Assuming mass of athlete is between 40 kg to 100 kg

/kkod dk nzO;eku 40 kg ls 100 kg ds e/; ekurs gq,here we will consider mass of athlete m = 50 kg

;gk ge /kkod dk nzO;eku m = 50 kg eku jgs

V = S/t =10

100= 10 m/sec

So, (vr%) K = 1/2 mv2 1/2 (50 102) = 2500 JSo Answer is (C)

vr% mkj (C) gS

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