Chapter # 8
Work and Energy
[1]
Objective - I 1.
Sol.
A heavy stone is thrown from a cliff of height h with a speed v. The stone will hit the ground with maximum speed if it is thrown (A) vertically downward (B) vertically upward (C) horizontally (D*) the speed does not depend on the initial direction. ,d Å¡ph pV~Vku ls ,d Hkkjh iRFkj v pky ls QSadk tkrk gSA iRFkj tehu ls vf/kdre pky ls Vdjk;sxk] ;fn bldks QSadk tk;s (A) m/okZ/kj uhps dh vksj (B) m/okZ/kj Åij dh vksj (C) {kSfrt fn'kk esa (D*) QSadus dh vkjfEHkd fn'kk esa pky fuHkZj ugha djrh gSA
D By energy canservation initial energy= final energy K.Bi + P.Bi = K.Bf + P.Bf 1/2 mu2 + mgh = 1/2 mv2 + 0
h
v = u 2 2gh So we can concluet that the speed does not depend on the initial direction. 2.
Two springs A and B (kA = 2kB) are strettched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is nks fLizaxksa A rFkk B (kA = 2kB) dks pkjksa fljksa ij leku ifjek.k ds cy yxkdkj izlkfjr fd;k tkrk gSA ;fn A esa lafpr ÅtkZ E gS rks B esa lafpr gksxh (A) E/2 (B*) 2E (C) E (D) E/4
Sol.
B Energy store in A = 1/2 kAxA2 = E ................ (1) SpringAand B stretched by applying equal magnitudes of force. For spring A kAxA = F ................ (2) For spring B kBxB = F ................ (3) From equation (2) & (3) kAxA = kBxB ................ (4) Given kA = 2 kB xB = 2xA Energy store in B = 1/2 kBxB2 =
1 kA (2x )2 = k x 2 A A A 2 2
= 2E (from equation (1) ). 3.
Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled out symmetrically to stretch the spring by a length x over its natrual length. The work done by the spring on each mass is fLizax fu;rkad k dh fLizax ds nksuksa fljksa ls nks leku nzO;eku tqMs+ gq, gSaA nksuksa nzO;ekuksa dks lefer :i ls [khapdj fLizax dks bldh ewy yEckbZ ls x nwjh ij izlkfjr fd;k tkrk gSA fLizax ds }kjk izR;sd nzO;eku ij fd;k x;k dk;Z gS (A)
Sol.
1 2 kx 2
(B) –
1 2 kx 2
(C)
1 2 kx 4
(D*) –
D 2
1 x 1 x Potential energy = k k 2 2 2 2 1 = kx2 4
m
2
1 2 kx 4
m k
m x/2
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m x/2
Chapter # 8 4.
Sol.
5.
Sol.
6.
Sol.
7.
Sol. 8.
Sol.
Work and Energy
[2]
The negative of the work done by the conservative internal forces on a system equals the change in (A) total energy (B) kinetic energy (C*) potential energy (D) none of these fdlh fudk; ij laj{kh vkarfjd cyksa ds }kjk fd;k x;k _.kkRed dk;Z fuEu ls rqY; gksrk gS (A) dqy ÅtkZ (B) xfrt ÅtkZ (C*) fLFkfrt ÅtkZ (D) buesa ls dksbZ ugha
C W = –v
(Change in potential energy)
The work done by the external forces on a system equals the change in (A*) total energy (B) kinetic energy (C) potential energy fdlh fudk; ij cká cyksa }kjk fd;k x;k dk;Z fuEu esa ifjorZu ds rqY; gksrk gS (A*) dqy ÅtkZ (B) xfrt ÅtkZ (C) fLFkfrt ÅtkZ
A Wexternal force = DE (Change in toatl energy) By work energy theorem Wext + Wint + Wpreudo + Wfriction + ............ = D K.E. Wint = – v Wext = K.B. + u Wext = T.E.
(D) none of these (D) buesa ls dksbZ ugha
The work done by all the forces (external and internal) on a system equals the change in (A) total energy (B*) kinetic energy (C) potential energy (D) none of these fdlh fudk; ij leLr cyksa ¼cká ,oa vkrafjd½ }kjk fd;k dk;Z fuEu esa ifjorZu ds rqY; gksrk gS (A) dqy ÅtkZ (B*) xfrt ÅtkZ (C) fLFkfrt ÅtkZ (D) buesa ls dksbZ ugha
B By work energy theorem
______________ of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is (A) kinetic energy (B) total mechanical energy (C*) potential energy (D) total energy f}d.k fudk; dh _________ dsoy d.kksa ds e/; nwjh ij fuHkZj djrh gSA mDr okD; esa fjDr LFkku dh iwfrZ ds fy;s lokZf/kd lgh p;u gS (A) xfrt ÅtkZ (B) dqy ;kaf=kd ÅtkZ (C*) fLFkfrt ÅtkZ (D) dqy ÅtkZ
C
A small block of mass m is kept on a rough inclined surface of inclination fixed in a elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be ,d fy¶V esa fLFkr vkufr dks.k okys [kqjnjs ur ry ij m nzO;eku dk NksVk CykWd j[kk gqvk gSA fy¶V Åij dh vksj ,d leku osx v ls xfr'khy gks tkrh gS rFkk ur ry ij CykWd uhps ugha fQlyrk gSA t le; esa ?kw.kZu cy }kjk CykWd ij fd;k x;k dk;Z gS (A) zero 'kwU; (B) mgvt cos2 (C*) mgvt sin2 (D) mgvt sin2
C Distance travelled by the elevator in 't' time is = vt Block don't slide on the wedge. So friction force = mg sin Work done by the force friction on the block in time t will be = F.d = Fd cos (90 - ) = mg sin . vt sin = mg vt sin
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90
m
f
v
Chapter # 8 9.
Sol. 10.
[3]
A block of mass m slides down a smooth vertical circular track. During the motion, the block is in (A) vertical equilibrium (B) horizontal equilibrium(C) radial equilibrium (D*) none of these ,d fpdus ,oa m/okZ/kj o`Ùkkdkj iFk ij m nzO;eku dk CykWd uhps dh vksj fQlyrk gSA bl xfr esa] CykWd gS (A) m/okZ/kj lkE; esa (B) {kSfrt lkE; esa (C) f=kT;h; lkE; esa (D*) buesa ls dksbZ ugha
D
A particle is rotated in a vertical circle by connecting it to a string of length and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is ,d d.k dks yEckbZ dh Mksjh ds ,d fljs ls cka/kdj m/okZ/kj ry esa bl izdkj ?kqek;k tkrk gS fd Mksjh dk nwljk fljk fLFkj jgrk gSA tc Mksjh {kSfrt gks rks d.k dh U;wure pky fdruh gksxh fd d.k o`Ùk iwjk dj ys (A)
Sol.
Work and Energy
g
(B)
2g
(C*)
3g
D Energy conservationAto B is T.Bi = T.Bf P.Bi + K.Bi = P.Ef + K.Bf 0 + 1/2 mu2 = mg (2) + 1/2 mv2 mu2 = 4mg + mv2 ........... (1) For minimum velocity 'u' that causes particle will complete the circle formation velocity tension at point 'B' is zero. F.B>D. of point 'B' mv 2 TB = 0 = mg From equation (1) & (2) mu2 = 4 mg + mg u=
5g
(D)
B
v
........... (2)
Am
5g
u
Objective - II 1.
Sol.
A heavy stone is thrown from a cliff of height h in a given direction. The speed with which it hits the ground (A*) must depend on the speed of projection (B*) must be larger than the speed of projection (C) must be independent of the speed of projection (D) may be smaller than the speed of projection h ÅapkbZ dh pV~Vku ls ,d Hkkjh iRFkj QSadk tkrk gSA ftl pky ls ;g tehu ls Vdjk,xk] og (A*) iz{ksi.k pky ij fuf'pr :i ls fuHkZj djrh gSA (B*) iz{ksi.k pky ij fuf'pr :i ls vf/kd gksrh gSA (C) iz{ksi.k pky ij fuf'pr :i ls fuHkZj ugha djrh gSA (D) iz{ksi.k pky ls de gks ldrh gSA
AB By apply the energy conservation T.Ei = T.Ef P.Bi + K.Bi = P.Bf + K.Bf mgh + 1/2 mu2 = 0 + 1/2 mv2
u 2gh Hitting velocity 'v' is depend on the speed of projection & large than the speed of projection. v=
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(Initial speed of stone is u) h
Chapter # 8 2.
Sol.
3.
Sol.
4.
Sol.
5.
Sol.
Work and Energy
[4]
The total work done on a particle is equal to the change in its kinetic energy (A*) always (B) only if the forces acting on it are conservative (C) only if gravitational force alone acts on it (D) only if elastic force alone acts on it. d.k ij fd;k x;k dqy dk;Z bldh xfrt ÅtkZ esa ifjorZu ds rqY; gksrk gS (A*) lnSo (B) dsoy rc ggh tc yxus okys cy laj{kh gksrs gSa (C) tc dsoy xq:Roh; cy gh bl ij yx jgk gksrk gSA (D) tc dsoy izR;kLFk cy gh bl ij yx jgk gksrk gS
A By work energy theorem The total work done an a particle is equal to the change in its kinetic energy.
A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. If follows that (A) its velocity is constant (B) its acceleration is constant (C*) its kinetic energy is constant (D*) it moves in a circular path d.k ij fu;r ifjek.k dk ,d cy bl izdkj yxrk gS fd bldh fn'kk lnSo d.k ds osx ds yEcor~ gksrh gSA bldk rkRi;Z gS fd (A) bldk osx fu;r gSA (B) bldk Roj.k fu;r gS (C*) bldh xfrt ÅtkZ fu;r gS (D*) ;g o`Ùkkdkj iFk ij xfr'khy gSA
CD Force of constant magnitude which is always pependicular to the to the velocity of the particle. That causes magnitude of velocity of the particle remain constant. So we can say that its kinetic energy is constant & this force is povide the centripetal acceleration that causes particle moves in a circular path. Consider two observers moving with respect to each other at a speed v along a straight line.They observe a block of mass m moving a distance on a rough surface. The following quantities will be same as observed by the two observers (A) kinetic energy of the block at time t (B) work done by friction (C) total work done on the block (D*) acceleration of the block. ,d ljy js[kk ds vuqfn'k nks izs{kd ,d nwljs ds lkis{k v pky ls xfr'khy gSA ;s izs{kd ,d m nzO;eku ds CykWd dks [kqjnjh lrg ij nwjh rd xfr djrk gqvk izsf{kr djrs gSaA nksuksa izs{kdksa }kjk fuEu jkf'k;ksa ds fy;s ,d leku jgsaxs (A) le; t ij CykWd dh xfrt ÅtkZ (B) ?k"kZ.k }kjk fd;k x;k dk;Z (C) CykWd ij fd;k x;k dqy dk;Z (D*) CykWd dk Roj.k
D Two observers moving with respect to each other at a speed v along a straight line mean with respect to each other change in kinetic energy is always same. By the work energy theorem we can say that work done by the friction force is same with respect to block is same & also say that acceleration will be same of the block. You lift a suitcase from the floor and keep it on a table. The work done by you on the suitcase does not depen on (A*) the path taken by the suitcase (B*) the time taken by you in doing so (C) the weight of the suitcase (D*) your weight. vki ,d lwVdsl dks Q'kZ ls mBkdj Vsfcy ij j[krs gSaA vkids }kjk lwVdsl ij fd;k x;k dk;Z fuEu ij fuHkZj ugha djrk gS (A*) lwVdsl }kjk r; fd;k x;k iFk (B*) vkids }kjk ;g dk;Z djus esa yxk le; (C) lwVdsl dk Hkkj (D*) vkidk Hkkj
ABD Work done by you on the suit case is change in Potential energy= mgh Here mg is the weight of the suit case & h is height of floor ro table.
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Chapter # 8 6.
Sol.
7.
Work and Energy
[5]
No work is done by a force on an object if (A*) the force is always perpendicular to its velocity (B) the force is always perpendicular to its acceleration (C*) the object is stationary but the point of application of the force moves on the object (D*) the object moves in such a way that the point of application of the force remains fixed. fdlh oLrq ij fdlh cy }kjk dksbZ dk;Z ugha fd;k tkrk gS] ;fn (A*) cy lnSo osx ds yEcor~ jgsA (B) cy lnSo blds Roj.k ds yEcor~ jgsA (C*) oLrq fLFkj jgs fdUrq cy dk fØ;k fcUnq oLrq ij xfr djsA (D*) oLrq bl izdkj xfr djs fd cy dk fØ;k fcUnq fu;r jgsA
ACD
Work done by a force on an object = F.d = Fd cos Displacement always in the direction of velocity. (a) F v mean = 90º, cos 90º = 0 Work done by a force in an object = 0 (b) Always acceleration not provide the direction of motion. (c) Object is stationary mean s d = 0 (d) It is the condition of the circular motion, here force provide the centripetal acceleration.Angle between this force & displacement is 90º. So work done by a force on an object is zero. A particle of mass m is attached to a light string of length , the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle (A*) the string becomes slack when the particle reaches its highest point (B) the velocity of the particle becomes zero at the highest point (C) the kinetic energy of the ball in initial position was
1 mv2 = mg. 2
(D*) the particle again passes through the initial position. ,d yEckbZ dh gYdh Mksjh ds ,d fljs ls m nzO;eku dk d.k tqMk+ gqvk gS] rFkk bldk nwljk fljk ca/kk gqvk gSA izkjEHk esa Mksjh dks {ksfrt j[kk tkrk gS rFkk d.k dks m/okZ/kj fn'kk esa v osx iznku fd;k tkrk gSA d.k o`Ùk iwjk ek=k dj ldrk gS (A*) tc d.k mPpre fcUnq ij igqapsxk rks Mksjh
(C) d.k dh izkjfEHkd xfrt ÅtkZ mv2 = mg FkhA Sol.
(D*) d.k iqu% izkjfEHkd fLFkfr ls xqtjsxkA
AD The minimum velocitygives to the particle is just able to complete a circle is 5g So initial K.E. of the particle = 1/2 mu2 = 1/2 m (5g) Apply energy conservationAto B T.BA = T.BB K.BA + P.BA = K.BA + P.BA 1/2 m (5g) + 0 = K.BB + 2mg K.BB = 1/2 mg 1/2 mv2 = 1/2 mg
B
Am
g End of complete one circle particle again passes through the initial position. vB =
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u
Chapter # 8 8.
Sol.
Work and Energy
[6]
The kinetic energy of a particle continuously increases with time (A) the resultant force on the particle must be parallel to the velocity at all instants. (B*) the resultant force on the particle must be at an angle less than 90° with the velocity all the time (C) its height above the ground level must continuously decrease (D*) the magnitude of its linear momentum is increasing continuously fdlh d.k dh xfrt ÅtkZ le; ds lkFk fujUrj c<+rh gS (A) d.k ij yxus okyk ifj.kkeh cy leLr {k.kksa ij osx ds lekUrj gSA (B*) gj le; d.k ij yxus okys ifj.kkeh cy dh fn'kk osx ds lkFk 90° ls de gSA (C) bldh tehu ls Å¡pkbZ fujUrj de gksuh pkfg,A (D*) blds js[kh; laosx dk ifjek.k fujUrj c<+ jgk gSA
BD Work done by the particle = K K = Fd cos F.d = K The resultant force on the particle must be at an angle less than 90º all the time. Linear momentum P = 2mK.E
9.
One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement , the work done by the spring is
1 kx2. The possible cases 2
are (A*) the spring was initially compressed by a distance x and was finally in its natural length (B*) it was initially stretched by a distance x and finally was in its natural length (C) it was initially in its natural length and finally in a compressed position (D) it was initially in its natural length and finally in a stretched position
,d gYdh fLizxa dk ,d fljk nhokj ls dlk gqvk gS rFkk nwljs fljs ls ,d CykWd tqMk+ gqvk gSA CykWd ,d {kSfrt ,oa fpduh lrg ij j[kk gqvk 1
gSA fLizax dk cy fu;rkad k gSA fdlh foLFkkiu esa fLizax }kjk fd;k x;k dk;Z 2 kx2 gSA laHko fLFkfr;k¡ gS -
Sol.
10.
(A*) izkjEHk esa fLizax x nwjh rd lEihfMr Fkh rFkk vafre fLFfr esa blus viuh izkÑfrd yEckbZ izkIr dj yh gSA (B*) ;g izkjEHk esa x nwjh rd izlkfjr Fkh rFkk vafre :i ls blus ewy yEckbZ izkIr dj yh gSA (C) izkjEHk esa bldh ewy yEckbZ Fkh rFkk vafre :i ls ;g lEihfMr fLFkfr esa gSA (D) izkjEHk esa ;g ewy yEckbZ esa Fkh rFkk vUr esa ;g [khaph gqbZ gSA
AB w = – u w = ui – uf The spring was initially conpress or streched by a distance x than ui = 1/2 kx2 Finallyin its natural length x=0 w = ui – uf = 1/2 kx2
K smooth
A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1s. (A) the tension in the string is Mg (B*) the tension in the string is F (C) the work done by the tension on the block is 20 J in the above 1 s. (D) the work done by the force of gravity is – 20 J in the above 1s. ,d nzO;ekughu rFkk ?k"kZ.k jfgr f?kjuh ls M nzO;eku dk ,d CykWd Mksjh dh lgk;rk ls yVdk;k x;k gS] Mksjh dk nwljk fljk ,d fu;r cy F }kjk [khapk tk jgk gSA ,d lsd.M esa CykWd dh xfrt ÅtkZ 20 twy c<+ tkrh gS (A) Mksjh esa ruko Mg gSA (B*) Mksjh esa ruko F gSA (C) mDr 1 lsd.M esa ruko }kjk CykWd ij fd;k x;k dk;Z 20 twy gSA (D) mDr 1 lsd.M esa CykWd ij xq:Rokd"kZ.k cy }kjk fd;k x;k dk;Z – 20 twy gSA
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Chapter # 8 Sol.
Work and Energy
B Tension in the spring is equal to F' T – mg = ma T = mg + ma Change in K = work done by the gravity Work done by the gravity = 20 J in 1 see Work done by the tension force = 0
[7]
T F a
m mg
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