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Notes On Work Enegry Power

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Published in: Physics
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Notes on Work Enegry Power for JEE aspirants.

Anil Y / Mumbai

9 years of teaching experience

Qualification: M.Tech (IIT Delhi (IITD), Delhi - 2012)

Teaches: Chemistry, KVPY Exam, Mathematics, Physics, AIEEE, CET, IIT JEE Advanced, IIT JEE Mains, Computer, Electrical, Electronics, Robotics, SAT, MBA Entrance

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  1. 5-Work, energy and power SYLLABUS Concept of work, energy and power. Energy : kinetic and potential. Conservation of energy and its applications, Elastic collisions in one and two dimensions. Different forms of energy. WORK It is defined as dot product of force and displacement W = P.s = F.s.cosO where is force acting on a point and S is displacement, 0 is the angle between force and displacement. w = f É.di Example 1: A block of ice is drawn through a distance 500 cm along a smooth horizontal surface. The pull in the rope in 100 dyne and the angle between the rope and ground in 300. Find the work done. Solution: Work done component of the force parallel to the displacement x displacement = F cos O x s = IOO cos 30 ox 500 = 43300 erg. Since displacement depends on reference frame, hence work also depends on the reference frame. Example 2. Solution: Power A block of mass 2kg is lying on a flat car, which is accelerating with constant acceleration of 1 m/s2. Find the work done by friction on the block in 10m journey of car (A) with respect to driver of car and (B) with respect to a ground observer. (A) With respect to driver, displacement of block is zero hence work done by friction force on block is zero. (B) With respect to ground observer, displacement is 10m and force of friction Work done by friction És.s 20J The rate at which work is done (by an agent) is known as power, which is delivered to the object by the agent. In other words, d 'VIV Power, P = dt since dW = F.dk dt where # is the force and di is the displacement in time dt. -s = the instantaneous velocity of the body) If we need to calculate average power then
  2. pav = Example 3. Solution: Net work done Time taken An engine of mass 20 tons pulls a train of 20 wagons, each of mass 20 tons with constant velocity of 72 km/hr on a level track. If the coefficient of kinetic friction is - 0.01, find the power developed by the engine. The total forward force F exerted by the k mg engine equals the frictional force ( = mg) on the whole train (including the engine), as the train is moving forward with constant mg velocity. The net power developed by the engine = F. v = (mg)v 72 x103 = 0.01 x (420 x 103) x 10 x = 420 kW. 3600 ENERGY Energy is the capacity (of a body or agent) to do work. Mechanical energy is of two types: Kinetic energy and Potential energy. KINETIC ENERGY It is the energy possessed by a body by virtue of its motion. If mass of body is m and speed v the kinetic energy K.E. = 1 2 —mv 2 POTENTIAL ENERGY Potential energy of a body or system is the energy possessed by the body by virtue of its position or strain. Potential energy function can only be defined for conservative force field only. In case of uniform gravity, potential energy is mgh where m is mass of body and h is height above the chosen reference level. Example 4. Solution: The potential energy of a system of two particles is given by U(x) = a/x2 — b/x. Find the minimum potential energy of the system, where x is the distance of separation; a, b are constants. a U(x) = 2 x b x dlJ (x) dx 2a b 3 2 x x d a dx b a 3 2 x x b x When the particle is in equilibrium 1 2 x
  3. 2b a Therefore, the minimum potential energy of the system is obtained by putting 2b a — in U(x) = — 2 a a 2 2b a x b 2b — a b x 3 a a 2 2 1 UNITS AND DIMENSIONS OF WORK, POWER AND ENERGY Work and Energy are measured in the same units. Power, being the rate at which work is done, is measured in a different unit. Quantity Units/Dimensions Dimension Absolute Gravitational Practical Other MKS FPS CGS MKS FPS CGS Work (Energy) M 1-2 T -2 Joule ft-Poundal erg kg-m ft-Cb gm-cm kWh, eV, cal Power M 1-2 T -3 Watt ft-poundal/sec Erg/sec. Kg-m/sec ft-Cb/sec. Gm-cm/sec HP, kW, MW Conversions between Different Systems of Units 1 Joule = 1 Newton x 1 m = 105 dyne x 102 cm = 107 erg 1 watt = 1 Joule/ sec = 107 erg/sec. 1 ft. poundal = 1 poundal x 1 ft = 13825 dyne x 30.48 cm Ikg-m = Ikg- wt x 1 m - "g" Newton x 1m = "g" Joule = 9.8 Joule 1 ft-lb = "g" poundal x 1 ft 32.2 poundal x Ift 32.2 ft-poundal 1 kWh = 103 watt x 1 hr = 103 watt x 3600 sec = 4.214 x 105 erg = 3.6 x 106 Joule 1 HP = 550 ft. Cb/sec (by definition) = 32.2 x 550 x 4.214 x 105 erg/sec = 746 watt. 1 MW = 106 watt. 1 cal = 1 calorie = 4.2 Joule lev = "e" Joule = 1.6 x 10-19 Joule . = 746 x 107 erg/sec. (e = magnitude of charge on the electron in colombs)
  4. POTENTIAL ENERGY OF SPRING Since spring force is conservative, hence potential energy function can be defined. dlJ = -É.di Since Fs = — kx IdlJ =fkx.dx U = —10
  5. m fvxdvx + fvydvy + fvzdvz — t or, by integration 2 — .dr .dF , if the motion takes place from A to B, we get or, or, or, m 2 v2A) = f Fnet.df 1 1 —mvA 2 wnet = AKE Conservation of Energy An interesting case of the work-energy theorem occurs when all the forces acting on a body are conservative. In this case, one can define a potential energy for each of these forces . fh.dF, IJ2= PI , ref fÉ2.dF , . etc • P 2 , ref Where PI, ref, P2,ref are the reference points for each force. The work -energy theorem can now be re-written, by using the relations, h.dF, dlJ2= É2.dF , dlJl = dlJl dlJ2 . mü. diü = m + CIUI + CIU2 + or, Integrating, as before, or, or, 1 2 —mv 2 1 2 —mv 2 1 —mVB + UI +1_J2 2 ......IBA = 0 + (UI +1_J2 + 1 —mv2A + (1$ 2 1 —mVA + UI + U 2 2 This result states that the total energy, 1 — mv2 + UI + IJ2 + ..... is conserved, when all the forces Etotal = Ekinetic + Epotential — 2 are conservative. MOTION IN A VERTICAL CIRCLE A particle of mass m is attached to a light and inextensible string. The other end of the string is fixed at O and the particle moves in a vertical circle of radius r equal to the length of the string as shown in the figure. Consider the particle when it is at the point P and the string makes an angle 0 with the downward vertical. Forces acting on the particle are T = tension in the string along its length mg = weight of the particle vertically downward. Hence net radial force on the particle is o 0 mgcos0 mgsin0
  6. FR = T — mg coso Since = circle where v = speed of the particle at P; R = radius of the Here R = / (length of the string) T mg cos0 = T = mg cose Since speed of the particle decreases with height. bottom, where cose = 1 (as 0 = 0) T max = mg at the top (4) Here v' = speed of the particle at the top. CRITICAL VELOCITY At the top, tension is given by 2 mg Hence tension is maximum at the where VT = speed of the particle at the top. (tangential) 2 For VT to be minimum, T z 0 VT = If VB be the critical velocity of the particle at the bottom, then from conservation of energy mg(2C) + —mv{ VT = 2mgC + —mgC = 5gc VB = 2 — mvB (6) Example 6. Solution : A heavy particle hanging from a fixed point by a light inextensible string of length I is projected horizontally with speed gC . Find the speed of the particle and the inclination of the string to the vertical at that instant when the tension in the string is equal to the weight of the particle. T mg coso o mgcose mgsin0 If v be the speed of the particle at B, then 2 ... (ii)
  7. From (i) and (ii), we get 2 T mg cos0 = ... (iii) Since at B, T = mg 2 mg(l — cos0) = v2 = gC (1 — cos0) ... (iv) Conserving the energy of the particle at point A and B, we have 1 2 mgc(l cos0)+ —mv2 —mv 2 Wherevo = gc and v = gc 1 -cose 2gC(1 - cos0) + gC (1 - cos0) cose = 2/3 Putting the value of cose in equation (iv) we get 3 1. COLLISIONS When two particles approach each other, their motion changes or their momentum changes due to their mutual interactions. This phenomenon is called collision. Generally, the collisions are of two types: (1) Elastic collision Elastic collision (2) Inelastic collision A collision is said to be elastic if kinetic energy is also conserved along with the linear momentum. There is no loss or transformation of kinetic energy. 21712 ml Different cases and 2rnl ml L.12 Case I: Case Il: If both bodies have the same mass, then ml = rn2=m In this case, 1.12 and 'v'2 = I-Il If one of the bodies say is initially at rest, then In this case, ml -m 2 2
  8. and 2ml ml +m2 2. energy are not conserved. mu -k '7721.12 and loss in kinetic energy E = I-Il -112 Inelastic collision, Those collision in which momentum are conserved, kinetic 2 milli + [7721.12 (ml + [772) 1 (ml + 1772, 2 or 222) 1 mu + mu 2 (Lil L.12)2 2 (ml +1772) Points to be Notice The maximum transfer of energy occurs if ml = (ii) If Ki and Kf are the initial and final kinetic energles of mass ml, decrease in its kinetic energy is given by 2 Further, if = nrnl and 112 = 0, then 4n 2. CONCEPT OF COEFFICIENT OF RESTITUTION the fractional When two bodies collide head—on, the ration of their relative velocity after collision and their relative velocity before collision is called the coefficient of restitution Thus e = Example 7: Solution: A moving particle of mass m, makes a head—on collision with a particle of mass 2m, which is initially at rest. Show that the colliding particle loses (8/9) of its energy after collision. Let u be the initial velocity of particle of mass m and v its velocity after the collision. Let V be the velocity of particle of mass 2m after the collision. From the principle of conservation of linear momentum, we have mu = mv+ (2m) V or The conservation of kinetic energy gives 1 — mu2= — mv2 +— (2m) V 2 or — b2 = 2 1/2 or (u—D) (U + D) = 21,2 Using Eq. (1) in Eq. (2) we have (2)
  9. Comparing (1) and (3) we get u or D = 3 Now, initial kinetic energy of the colliding mass is K = — mu2 2 Final kinetic energy, Loss in kinetic energy is AK = — mu2— — rnv2 2 2 AK . Fractional loss = (3) (4) 1 2 mu 2 1 1 2 mu 2 2 mu 2 2 2 u 2 u 2 1 3 8 9 Example 8: Solution : OBJECTIVE A neutron moving at a speed v undergoes a head-on elastic collision with a nucleus of mass number A at rest. Prove that the ratio of the kinetic energies æA-1ö of the neutron after and before collision is 8A+1Ö Mass of neutron (ml) = 1 unit. Mass of nucleus (rn2) = A units. Here I-Il = u and 1.12 = 0. Therefore the velocity of the neutron after the collision is - m2ö æl- A ö 81 + A O KE of neutron after collision 1 —mlV12 2 1 1 2 1 2 — mil-I 2 æI-Aö 2 81+AÖ 1 1 2 2 KE of neutron before collision æ1-Aö 81+AÖ Their ratio is
  10. Solution: 2: Solution: Solution : 4: Solution : A block of mass 10 kg accelerates uniformly from rest to a speed of 2 m/s in 20 sec. The average power developed in time interval of O to 20 sec is (A) IOW (C) 20W (B)Average power Net work done pav = Total time taken Net work done = change in kinetic energy = Final energy — initial energy 1 - = 20J 2 Average power - - = 1 watt 20 An object is attached to a vertical spring and is allowed to fall under the gravity. What is the distance traversed by the object before being stopped? (A) mg/k (C) mg/2k (B) 2mg/k (D) none of these. 1 - — kx2 x = 2mg/k (B)Apply COE; mgx- 2 When a man walks on a horizontal surface with constant velocity work done by the (A) frictional force is zero (C) gravity is zero (B) contact force is zero (D) man is zero Since mg & N are perpendicular to velocity and dé , work done by these forces is zero. Since no relative sliding occurs during walking, static friction comes into play. Hence the point of application of static frictional force does not move. Therefore no work is done by frictional force. ds mg Man has to lose his body's (internal) energy E, hence performs work because W = AE (numerically). (A) A pumping machine pumps water at a rate of 60 cc per minute at a pressure of 1.5 atm. The power delivered by the machine is (A) 9 watt (B) 6 watt (D) None of these (D)Power = F.v Where F = force imparted by the machine , v = velocity of the liquid P = p.A.v, Where p = pressure & A = effective area dV dt 3 60 -6 2 60 1 atm 105 N/m2 ) = 0.15 w none of these A block of mass m moving on a smooth horizontal floor, with a constant velocity vo collides with a light spring of stiffness constant k which is
  11. rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the spring on the block is F, then: (D) all of these Solution : Solution : Solution : (C)Energy conservation between the positions A and B 1 —mvä APE+ AKE = O — 2 2 m k Fmax = kx = km vo rum TVO. vo k A block of mass m is moving with a constant acceleration a on a frictional plane. If the coefficient of friction between the block and ground is u, the power delivered by the external agent after a time t from the beginning is equal to: (A) ma2t (B) umgat (D)lnstantaneous power delivered where, Fv ma f p (f ma) put P = (gmg + ma)v = m(a + ug).at Find the ratio of work done by gravity on the block under the two conditions. Assume = O between block and plane. pcmg h Condition (1) h 02 Condition (2) (C)Work done by gravity will depend only upon the height 'h' and not on the length of inclined plane. So work done in both the vases = mgh only. ratio = 1 A spring, placed horizontally on a rough surface is compressed by a block of mass m, placed on the same surface so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is V, the potential energy stored in the spring is g2rn2g2 (A) 2k Vt2m2g 2k 2gm g (B) k 3kt2mg 2 (D) k
  12. 'Solution : 9: Solution : Solution : 11: Solution : 12: Solution : 13: (A)For equilibrium of the block F max - = O gmg 2k Y2m2g2 2k An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts with zero magnitude and gradually increases until the string makes an angle 9 with the vertical. Work done by the variable force F is: (A) mgL(1-sin9) (C) mgL(1-cos9) (B) mgL (D) mgL(1+cos9) Work done by the external force = A E of the system (object) = APE e: AKE = mgh = mg (L — L cos 0) = mgL (1 —cos 0). L cos 0 h When a body of mass M slides down an inclined plane of inclination 9, having coefficient of friction through a distance s, the work done against friction is: (A) Mg cos 0 s (C) Mg (PI cos 0 - sin O)s (A)Work done (B) Mg sin Os (D) None of the above against - work done by friction = pcmg cos 0 s. An object is pushed and pulled along a certain angle w.r.t. horizontal and is moved the some distance on the ground. Now the work done by friction pull push frictional force will be (A)push effect. (C) remain same. insufficient. (D)fr = Pi(mg F sinu) pulling. Fv = Pi(mg + F sinu) pushing. (B)pull effect (D) data f sinu f cosoc A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to: (D) None of the above s (D)W = f F.ds= f-ds = kin (S/SI) A particle at rest is constrained is to move on a smooth horizontal surface. Another identical particle hits the stationary particle with a velocity v at an angle 0 = 600 with horizontal. If the particles move together, the velocity of the combination just after the impact is equal to
  13. Solution : 14: Solution : (D)mv cos 0 = (M + m) v' — cos 600 2 4 m Find the horizontal velocity of the particle when it reach the point Q. Assume the block to be frictionless. Take g = 9.8 m/s2. (A) 4 m/s (C) 3.13 m/s m=2kg P 1 (C)mg 1 2 1 2 2 (D) 3.6 m/s 5m 2 3.13 m/s. 2 0.5m 15. sol: 16. sol: 17. A neutron mass 1.67 x 10-27 kg moving with a velocity 1.2 x 107 ms-I collides head- on with a deuteron of mass 3.34 x 10-27 kg initially at rest. If the collision is perfectly inelastic, the speed of the composite particle will be (A) 2 x 106 ms-I 4 x 106 ms-I (B) (C) 6 x 106 ms-I (D) 8 x 106 ms-I (B) Mass of neutron (m) = 1.67 x 10-27 kg. Speed of neutron (v) = 1.2 x 107 ms-I. Notice that the mass of deuteron (M) = 3.34 x 10-27 kg = 2m. If V is the speed of the composite particle, the law of conservation of momentum gives mv = (m + mv mv or or m +M m+ 2m v 1.2' 107 4' 106 ms 3 3 In above. Q. if the collision were perfectly elastic, what would be the speed of deuteron after the collision ? (A) 2 x 106 ms-I (C) 6 x 106 ms-I 4 x 106 ms-I (B) 8 x 106 ms-I (D) (D) In an elastic collision, both momentum and energy are conserved. Using the two laws, it is easy to see that ( M = 2m), the deuteron will move forward with a speed æ 1.2' 107 0.8 107 ms 2v/3 3 A body of mass 5kg initially at rest, is subjected to a force of 40N. Find the kinetic energy acquired by the body at the end of 5 seconds (A) 4000 J (C) 6249 J (B) 6000 J (D) 6145 N Solution: (A):. Here we will calculate acceleration
  14. F - = 8 m/s2 Velocity at the end of 5 seconds is v = u + at v = o + = 40 m/s K.E. acquired = —mv2= 1 2 18. A body is constrained to move along z-axis of the co-ordinate system is being applied by a constant force (27 + + 4k). Find the work done by this force in moving the body over a distance of 5m along z-axis. (A) 30 J (C) 40 J (B) 20 J (D) 61 J Solution: (B)Here displacement is only along z-axis 19. Hence, é=5k and 27+3j+4k Work done by force W = É.é A body moves from point A to point B under the action of a force varying in magnitude as shown in the force displacement graph. Find total work 10 5 done by the force (A)50J (C) 40 J Solution: (D)Work done dw = .dS 5 (B) 60 J (D) 45 J w = f É.dé = Area under F-S curve = 1 -x 5x 45J 2 20. Solution : 21. A body dropped forma height 'H' reaches the ground with a speed of 1.1 gH . Calculate the work done by air friction? (A) 0.395 mgH (C) 0.400 mgH (B) -0.395 mgH (D) -0.400 mgH 1 - .21)gh = 0.605mgH 2 Now AKE = mgH fralr 0.605 mgH = mgH — fralr = 0.395 mgH A block of mass m = 2kg is attached to two unscratched springs of force constant kl = 100 N/m and = 125 N/m. (D) AK.E. of particle on reaching the ground o f = air friction mg Now since body moves opposite to the direction of force of air friction. o(air friction) = 0.395 mgH 2kg 125 N/m 100 N/m 0.1m
  15. The block is displaced towards left through a distance of 10 cm and released. Find the speed of the block as it passes through the mean position. (A) 1.06 m/s (C) I .04 m/s (B) 1.02 m/s (D) 1.05 m/s Solution: (A)Here effective spring constant of the combination = kl + k2 = Now applying the energy conservation principle, we have 1 1 2 — X (0.1 — mv 2 2 1 1 — X (0.1 2 2 v2 = 1.125 * v = 1.06 m/s at the mean position. 225 N/m 22. A 16kg block moving on a frictionless horizontal surface with a velocity of 4m/s compresses an ideal spring and comes to rest. If the force constant of the spring is IOON/m, then how much is the spring compressed? (A) 1.7m (C) 1.4m Solution: (B)Here energy is conserved. (B) 1.6m (D) 1.8m 1 -10
  16. (A) 16.825 N-m/s (C) 16.852 N-m/s (B) 16.822 N-m/s (D) 16.528 N-m/s Solution : (A)The total downward force acting on the block 2 2 = 5[0.5 + 0.173] = 3.365 N. Now the power required to move up along the inclined power = 3.365 x 5 = 16.825 N-m/s mg sino mg cose 25. Solution: 26. 27. A small body of mass m is located on a horizontal plane. The body acquires a horizontal velocity vo. Find mean power developed by the frictional force, during the whole time of its motion. If coefficient friction 0.27, mass of body m = Ikg, and Vo = 1.5 m/s (A) gungvo gungvo 4 gmgv (B) 2 gungvo (D) 8 (B) Work done against friction w = - umgs dw dt pa v = o ds — umgs. dt um g — gmgs.v vo v.dv = gmgv 2 A block of mass m moving with a velocity vo on a smooth horizontal floor collides with a light spring of stiffness k that is rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the spring on the block is F, then prove that F = km (A) km vo (C) 2 km vo (B) 4 km vo Solution : (A)App/ying work energy theorem APE+ AKE = O 1 1 —mvo 2 2 m k Fmax = kx = km vo A bus of mass m produces a constant power P. If the resistance to motion is R. Find the maximum speed at which the bus can travel on level road and acceleration of bus when it is travelling at half of its maximum speed. m
  17. 1 2m m Ans. (B) Solution : Speed will be maximum, when engine of bus will produce maximum power i.e. p p P = R .Vmax Vmax Here v = max 2 Force by which engine pulls the bus is max 2 p Since F —R = m.a 2 R-R=m.a m A uniform rod of length L and mass m hinged at one end is hanging 28. vertically. The other end is now raised until it makes an angle 600 with vertical. How much work is required? (A) mgC mgt 6 Ans. (D) mgt (B) 8 4 Solution: We can consider its whole mass is centre ted at its centre of mass O. work done = mgh c c — —cos600 = mg 2 1 = mgC 2 600 2 1 4 mgt 4 29- 32. 29. A point mass m starts from rest and slides down the surface of a frictionless solid sphere of radius r as shown in figure. Measure angles from the vertical and potential energy from the top. Find The change in potential energy of the mass with angle mgR(1 -cos0) (B)AU = + mgR(1 -cos0) (C)AU = + mgR(1 + cos0) mgR(1 + cos0) Ans. (A)
  18. Solution:Consider the mass when it is at the point B. I-JA (P.E. at A) = O UB(P.E. at B) = mgR (1 -cos0) Al-J= mgR(1 -cos0) 30. Negative sign indicates that P.E. decreases as particle slides down. The kinetic energy as a function of angle (A) T = mgR(1 - (B) T = mgR(1 + cos0) (C) T mgR(1 -cos0) -cos0) Ans. (A) Solution:Conserving energy at points A and B. IJA + UB + TB where I-JA = P.E. at A, IJB= P.E. at B K.E. at A, K.E. at B + = mgR(1— COSO) + TB T = mgR(1 — coso) 31. The radial and tangential acceleration as a function of angle. (A) at = sino Ans. (B) Solution:Since T = —mv2 2 1 (B) at = gsino = —gsin0 32. = mgR(1 - cos0) =D2mgR sin20/2 2 2 v = 2 gR sinO/2 aradial = V2/R ar = 4g sin20/2 at = g sino also, in circular motion velocity is along the tangent, therefore Vtangential = 2 gR sinO/2 at = dt at = Al(gR) COS0/2 = M(gR) ocos0/2 at = W(g/R) v cos0/2 , as OR = v at = gsin0 The angle at which the mass flies off the sphere. (A) o = cos-l(4/3) 1(1/3) (C) o = cos-l(2/3) I (2/8) Ans. (C) (B)O = cos- (D)O = cos- mgcos t - cos9) mgsin9
  19. Solution : For circular motion mg cose - N = at the moment when the particle brakes off the sphere N = 0. mg cos0 = g cose = v2/R v = 2 gR sin0/2 g cose = 4g sin20/2 = 2g(1 - cos0) cose = 2/3 O = cos-l(2/3) A block of ice mass 10 kg slides down an incline 5 m long and 3 m high. 33-37. A man pushes up on the ice block parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and the incline is 0.1. Find : The work done by the man on the block 33. (B) = -260 J (D) Wm = -560 J Solution : 34. (A) = -300 J (C) = -400 J Ans. (B) From the figure sino - - 3/5 4/5 F.B.D. of the block and cose = 5m 3m mgsin0 F = force by the man f = frictional force mgcos0 N = normal reaction of the surface mg = gravity Since block slides with constant speed, hence mg sin0D= F + f F = mg sino -f 3 4 5 5 as f = PI mg cose man = F.S = FScos180 = -FS Here F -52 N, and S = 5m. -52 x 5 J = -260 J 52 N, The work done by gravity on the block (A) Wgravity = 300 J (B) Wgravity = 400 J
  20. (C) Wgravity = 600 J Ans. (A) (D) Wgravity = 500 J Solution : 35. Solution : 36. Solution : 37. Solution : 38: Solution: 39: Wgravity = mg S sinO= 10 x 10 x (3/5) J = 300 J The work done by the surface on the block (A) = + 40 J (C) Wsurface = -41 J Ans. (D) (B) wsurface= - 42 J (D) Wsurface = 40 J Wsurface = WN + Wfriction = 0 + fS cos1800 = 0 - pcmg cose S = -0.1 (4/5) x 5 J = -40J The work done by the resultant forces on the block Ans. (A) Work done by the resultant force is given by W = + wg + WN + = -260 J + O The change in K.E. of the block . Ans. (A) Since net work done by all the force is zero, hence change in K.E. = 0 If a constant force F = 3i+ 4 j+5k is acting on a particle and displacement of particle becomes r = 3i+ 4 j+6k. Find net work done by force F. (A) 55 J (C) 90 J (B) 80 J (D) 100 J (A) w (3 +4j +4j +6k) 9+16+30 = 55 J A particle is moving on circle of radius 5mt. A force of constant magnitude 10 N is acting on particle along tangent find work done by this force when particle will complete one circle. (A) 1 OOTJ (B) 200TJ (C) 300ßJ (D) 400fiJ Work is zero because displacement is zero. Correct solution: (A) Force F is varying because its direction is changing so w F.r dw = Fdr - Fdr where dr is a small path length and small path length will always be along tangent fFdr Ffdr Fx2nr 1 001tJ dr
  21. 40: Solution: A block of mass 10kg is pulled by a force F having magnitude 20N. Find work done by force F if body moves 5mt in right direction. Given that 10 = 1 mt. and x = 25mt initially. In this case force F is variable Common mistake: F— Fcos0 i+Fsin0j dw — F.dxi = [F cos0i+ F sino j]. dx i = F cos0 dx x cos0= 102 + F 102 + x 102 + 25 We know that in this case angle between F and displacement is acute so work should be positive, contradiction has come. text Correct solution: (B) y = 25 — x dy dw = (F + F sinoj). dy I dw = —Fcos0 dx 20 102 + 25 w = — dx and 20 41: Solution: A projectile is projected with initial velocity 20m/s and at an angle of 300 from horizontal. Find the total work done when it will hit the ground. (B) 10 (C) 20 (D) 30 (A)work = (mg j) . (r i) = 0 20 m/s 300 42. A particle of mass 1 kg is moving along x-axis and a force F is also acting along x-axis and a force F is also acting along x-axis in such a way such that its displacement is varying as: - x = 38. Find work done by force F when it will move 2mt. (A) 12 J (C) 32 J (B) 16 J (D) 42 J
  22. Solution: (A) x 43: Solution:(D) = 3t2 dx dt Acceleration 6t m/ s dv 6m /s2 dt Force acting on particle = ma - -Ix6=6N Displacement = 2 work = work F.r 12J 2 If a projectile is projected with initial speed u and angle 0 from horizontal then what will be its average power up to time when it will hit the ground again. (A) 100 (C) 300 20 m/s 300 (B) 200 44: When it will hit the ground the net work done is zero. So p 0 If a particle is moving on straight line and a constant instantaneous power is supplying on the particle then find displacement of particle as a function of time. 3/2 1/3 1 3 5 3 2 3 m m m 1/3 1/3 1/3 3/2 3/2 3/2 10 3C Solution: (C) Fv = constant = c
  23. mAV = c c c mv dv c —JEdx dx mv dx dt 3 c m 1/3 1/3 — Ex m 3/2 1/3 1/3 45. Solution: 46. -2 Solution: Solution : A fast moving neutron suffers one-dimensional elastic collision with a nucleus 7 Ni4 . Wha approximate percentage of energy is lost by the neutron in the collision ? (B) 100/0 5600 (C) 24.9 - 25 (1+14)2 225 An 8 kg block accelerates uniformly from rest to a velocity of 4 ms-I in 40 second. The instantaneous power at the end of 8 second is (A) 0.64 w (c) 0.16 w -2 ms 40 (B) 0.32 W (D) 0.08 w. 1 —ms 10 Velocity at the end of 8 second = 0.8ms 10 1 Required power = (ma) v = 8 8=0.8 w=o.64W 10 Work is which type of physical quantity. (A) vector (C) tenser (B) (B) scalar (D) None of these. 48: An engine is working at a constant power draws a load of mass m against a resistance r. Find maximum speed of load and time taken to attain half this speed. pm 8r2 pm 2 pm 8r pm 9r2 Solution: (A)
  24. At the time of maximum velocity f = r I e net force onload-0 V max max dv p dt v p.2r v.dv pm 8r2 t dt 49. Block m has given a velocity vo towardly right. What will be the maximum power delivered by the spring force during the entire motion. 511 40 37t 40 In the given case v = vo COS(ot) sinot Acceleration = - vo o sin(ot) smooth Solution: Force = mass x acceleration = mvoo sin(ot) Power = - force v = mv020 sin(ot) cos (ot) CD sin(20t) t max power mv max 50: mvoo for t = 37t So Ans = — 40 It 37t 40 311 70 lit 40 minimum power A particle of mass 10kg is moving with velocity
  25. 2i + 3j + 4k what will be its kinetic energy. If A constant force F = i - j - 2k starts acting on body then find after 2 second its power and work done by this force in 2 second. 10 - Solution: 74 105 Acceleration = mass S at + — at2 (2 + 3 j + 4k) x 2 -k — x 42 x 84 84 (D) m/s2 10 j -2k 21 29 38 21 Work = F.S= — 5 4+9+16 84 Ak.E 84 v: -29 25 641 m/s 10 84 29 76 5 29) (D) t(sec) smooth If a spring mass system is placed on a horizontal smooth surface. Then what will be the net work done by spring force after one time period. (A) —km2 -—km2 Solution: -—km2 (B) (D) +-km2 0 —mv + CD spring friction —kx2 — gmgx —mv > + 2gmgx mvt 0 2grng+ 4k12m2g2 + 4kmv02 2k pcmg+ P12m2g2 + 4kmv{
  26. k 1 —km2 So work by spring = 2 k (B) Check whether the force F = xi+ yj + zk is conservative or not. Solution: 53: Solution: (A) conservative (B) non conservative (C) variable (D) none of these. Let a particle on which force F is acting moving from (Xl, Yl, ti) to (x2, Y2, t2) then dw = F.dr and dr = dxi + dyj + dzk dw = (xi + yj = zk).(dxi + dyj + dzk) (x2,Y24) 2 2 2 x z 2 22 (X! ,YIZI ) So above work depend only on the coordinates of point (Xl, Yl, Zl) and (x2, Y2, u). So force is conservative. (A) Prove that the central force which follows the inverse square law are conservative forces. 1 1 1 1 (D) k(rl k 1 x Let consider a central force F . Now we will 2 prove that if a body on which the force is acting F B then work done on it will not depend on both F r = xi+Yj> dr =dxi+dyj k low dw [xdx+Ydy] 2 3/2 k will move from A to k(xi + Yj) 2 3/2 (XI,YI) (x 2 + Y ) 1 k 1 + y 12 1 1 k (B) 54: Prove that # = 2 x j+ z2 k is a conservative force.
  27. X X Xl + Y2 Xl + Y2 x 12 + x 12 + Y2 Y12 + Y12 + y 12 + y 12 + z z 10 Solution: 55: Solution 56: F = + y 2 j + z2k Let particle moves from (Xl, Yl, Zl) to (x2, )/2, u) dw = x2 dx + y2dy + z2dz ( C —xi + É —yi + z22 —zi which depend only on (Xl, Yl, Zl) and (x2, Y2, u) So F will be conservative. (A) A particle of mass m is moving in a circular path of constant radius r such that centripetal acceleration ac is varring with time t as k2r t2 where k is constant what is the power delivered to the particle by the forces acting on it. mk4r2t5 (A) mk4r2t5 dv k2rt2 rt3 mk4r2t5 (B) mk4r2t5 (D) Centripetal force will not supply the power and power by the tangential dv force = m dt k2rt3 = mk2rt2 mk4r2t5 Find the total compression in the spring. m 1/02 g ax Smooth amgL2
  28. mV02 — amgL2 k m 1/02 m V02 amgL2 amgL2 k 2 Solution: 57. Solution: 58: Solution: 59: Solution: • = ETf — work done by nonconservative force 1 —mv: 2 1 E —kx2 2 amgL2 ax mg dx 2 — ml/' —kx2 + amgL2 1 2 2 2 mV02 — amgL2 (A) k Power applied to a particle varies with time as P = (4t3— 5t + 2)watt, where t is in second. Find the change is its K.E. betwæn time t = 2 and t = 4 sec. (A) 212 J (C) 214 J p = ma V dV — = 4t3 = mv dt 2 2 (B) 213 J (D) 215 J 5t2 — + 2t 2 So change in K.E. will be 214 J. (C) An engine develops 10KW of power. How much time will it take to lift a mass of 200kg to a height 40m. (g = 10 ms-2) (A) (C) 8s w p W mgh p p 10x100 (D) IOS What is work done when body will displace perpendicular to the force. (A) zero (C) minimum (A) F.d = Fd cose 900 (B) maximum (D) None of these.
  29. Solution: then W = O (A) A tube-well pump out 2400kg of water per minute. If water is coming out with a velocity of 3m/s, the power of the pump is (A) 120W (B) 180 W (C) 240 w (D) 90 W (B) Mass of water pumped per second 2400 40 kg m 60 v = 3m/ s Kinetic energy of water coming out per second 1 2 —mv 2 1 2 Power of pump = 180 J /s = 180 W (B)

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