Physics-13.5-Work, energy and power

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 = = F.s.cos where is force acting on a point and is displacement,  is the angle between force and displacement. W = 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 30°. Find the work done. Solution : Work done = component of the force parallel to the displacement  displacement = F cos   s = 100 cos 30 ° 500 = 43300 erg.  Since displacement depends on reference frame, hence work also depends on the reference frame. Example 2. 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. Solution: (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 fs = ma = 2  1 = 2N  Work done by friction . = 2  10 = 20J Power 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, Power, P = = , since dW = where is the force and d is the displacement in time dt. = ( = = the instantaneous velocity of the body) If we need to calculate average power then Pav = Example 3. 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 k = 0.01, find the power developed by the engine. Solution: The total forward force F exerted by the engine equals the frictional force ( = k mg) on the whole train (including the engine), as the train is moving forward with constant velocity. The net power developed by the engine = F. v = k (mg)v = 0.01  (420  103)  10  = 420 kW. 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. = 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. 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. Solution: U(x) =  F =  F =   F = When the particle is in equilibrium F = 0  2/x3  = 0  x = Therefore, the minimum potential energy of the system is obtained by putting x = in U(x) = , Umin = = . 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  Work (Energy) Power Dimension ML2T-2 ML2T-3 Absolute MKS Joule Watt FPS ft-Poundal ft-poundal/sec CGS erg Erg/sec. Gravitational MKS kg-m Kg-m/sec FPS ft-b ft-b/sec. CGS gm-cm Gm-cm/sec Practical (Other) kwh, eV, cal HP, kW, MW Conversions between Different Systems of Units 1 Joule = 1 Newton  1 m = 105 dyne  102 cm = 107 erg 1 watt = 1 Joule/ sec = 107 erg/sec. 1 ft. poundal = 1 poundal  1 ft = 13825 dyne  30.48 cm = 4.214  105 erg 1kg-m = 1kg- wt  1 m = "g" Newton  1m = "g" Joule = 9.8 Joule 1ft-lb = "g" poundal  1 ft 32.2 poundal  1ft 32.2 ft-poundal 1 kwh = 103 watt  1 hr = 103 watt  3600 sec = 3.6  106 Joule 1HP = 550 ft. b/sec (by definition) = 32.2  550  4.214  105 erg/sec. = 746  107 erg/sec. = 746 watt. 1 MW = 106 watt. 1 cal = 1 calorie = 4.2 Joule 1eV = "e" Joule = 1.6  10-19 Joule (e = magnitude of charge on the electron in colombs) POTENTIAL ENERGY OF SPRING Since spring force is conservative, hence potential energy function can be defined. dU = –  U = = – Since Fs = – kx  = U = This is the potential energy stored in the spring of force constant k. Example 5. Find the maximum energy stored in the spring shown in the figure, for which the block remains stationary on the rough horizontal surface. Solution: Let the spring be compressed by x. The potential energy stored in the spring = U = 1/2 Kx2 for maximum potential energy in the spring. N – Mg = 0  N = mg Therefore, fmax = mg kxmax = mg  xmax = The maximum potential energy stored in the spring = 1/2 kxmax2 = = . WORK KINETIC ENERGY THEOREM It is possible to relate the work done by all the forces on a body (or a system) to the change in kinetic energy of the body (or, the system). Consider a rigid body of acted upon by forces , . . . moving with a velocity which is in general a function of time. Newton's second law gives us : m . . . (1) where RHS represents the resultant (net) of all the forces , . . . . .etc. Taking the dot product of both sides of equation (1) by dt (=d , since ), we get, m . dt = + . . . . = . d or, m = + . . . . (2) Now, , and = dvx + dvy + dvz ; so, m (vx dvx + vydvy + vzdvz) = .d or, m = , if the motion takes place from A to B, we get by integration or, m or, or, m . . . (3) Wnet = KE 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 : U1 =  , U2 =  , . . etc. Where P1, ref, P2,ref are the reference points for each force. The work -energy theorem can now be re-written, by using the relations, dU1 =  , dU2 =  , . . . m . =  dU1  dU2 . . . or, m + dU1 + dU2 + . . .. = 0 Integrating, as before, + = 0 or, or, This result states that the total energy, Etotal = Ekinetic + Epotential = mv2 + U1 + U2 + . . . . . is conserved, when all the forces 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  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 FR = T – mg cos . . . (1) Since FR = , where v = speed of the particle at P; R = radius of the circle Here R = l (length of the string)  T  mg cos =  T = + mg cos . . . (2) Since speed of the particle decreases with height. Hence tension is maximum at the bottom, where cos = 1 (as  = 0)  Tmax = . . . (3) Tmin = at the top (4) Here v = speed of the particle at the top. CRITICAL VELOCITY At the top, tension is given by T = where vT = speed of the particle at the top. (tangential)  For vT to be minimum, T  0  vT = . . . (5) If vB be the critical velocity of the particle at the bottom, then from conservation of energy mg(2) +  vT =  2mg + mg =  vB = (6) Example 6. A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed . 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. Solution : FR = T  mg cos . . . (i) If v be the speed of the particle at B, then FR = . . . (ii) From (i) and (ii), we get T  mg cos = . . . (iii) Since at B, T = mg  mg(1  cos) =  v2 = g (1  cos) . . . (iv) Conserving the energy of the particle at point A and B, we have Where vo = and v =  g= 2g(1 - cos) + g (1 - cos  cos= 2/3 . . . (v) Putting the value of cos in equation (iv) we get v = 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 (2) Inelastic collision 1. Elastic 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. and Different cases Case I: If both bodies have the same mass, then m1 = m2 = m In this case, v1= u2 and v2 = u1 Case II: If one of the bodies say m2 is initially at rest, then u2 = 0 In this case, and 2. Inelastic collision, Those collision in which momentum are conserved, kinetic energy are not conserved. and loss in kinetic energy or Points to be Notice (i) The maximum transfer of energy occurs if m1 = m2, (ii) If Ki and Kf are the initial and final kinetic energles of mass m1, the fractional decrease in its kinetic energy is given by Further, if m2 = nm1 and u2 = 0, then 2. CONCEPT OF COEFFICIENT OF RESTITUTION 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 Example 7: 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. Solution: Let u be the initial velocity of particle of mass m and  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 u –  = 2 V … (1) The conservation of kinetic energy gives mu2 = mv2 + (2m)V2 or u2 – v2 = 2V2 … (2) or (u – ) (u + ) = 2V2 Using Eq. (1) in Eq. (2) we have 2V (u + ) = 2V2 or u +  = V or 2(u + ) = 2V … (3) Comparing (1) and (3) we get u –  = 2 (u + ) or  = … (4) Now, initial kinetic energy of the colliding mass is Ki = mu2 Final kinetic energy, Kf = m2 Loss in kinetic energy is K = Ki – Ki = mu2 – m2  Fractional loss = = = = 1 – = 1 – (  = – u/3) = Example 8: 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 of the neutron after and before collision is . Solution : Mass of neutron (m1) = 1 unit. Mass of nucleus (m2) = A units. Here u1 = u and u2 = 0. Therefore the velocity of the neutron after the collision is KE of neutron after collision KE of neutron before collision . Their ratio is . OBJECTIVE 1: 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 0 to 20 sec is (A) 10W (B) 1W (C) 20W (D) 2W Solution: (B)Average power Pav = Net work done = change in kinetic energy = Final energy – initial energy = = 20J . . . (1) Average power = = 1 watt 2: 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 (B) 2mg/k (C) mg/2k (D) none of these. Solution: (B)Apply COE; mgx= kx2  x = 2mg/k 3: When a man walks on a horizontal surface with constant velocity work done by the (A) frictional force is zero (B) contact force is zero (C) gravity is zero (D) man is zero Solution : Since mg & N are perpendicular to velocity and , 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. Man has to lose his body's (internal) energy E, hence performs work because W = E (numerically). (A) 4: 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 (C ) 9 kW (D) None of these Solution : (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 = p = ( 1 atm  105 N/m2 ) = 0.15 w  none of these 5: A block of mass m moving on a smooth horizontal floor, with a constant velocity v0 collides with a light spring of stiffness constant k which is rigidly fixed horizontally with a vertical wall. If the maximum force imparted by the spring on the block is F, then: (A) F  (B) F (C) F  v0 (D) all of these Solution : (C)Energy conservation between the positions A and B PE+ KE = 0  kx2  m =0  x =  Fmax = kx =  Fmax  k m v0. 6 : 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 , the power delivered by the external agent after a time t from the beginning is equal to: (A) ma2t (B) mgat (C) m(a+g)gt (D) m(a+g)at Solution : (D)Instantaneous power delivered = P = = Fv where, F – f = ma  F = f + ma  P = (f + ma) v Put f = mg  P = (mg + ma)v = m(a + g).at 7: Find the ratio of work done by gravity on the block under the two conditions. Assume  = 0 between block and plane. (A) 1 : 2 (B) 2 : 1 (C) 1 : 1 (D)  :  Solution : (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 : 1 8: 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 , the potential energy stored in the spring is (A) (B) (C) (D) `Solution : (A)For equilibrium of the block Fmax - mg = 0  F = mg  U = 9: 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  with the vertical. Work done by the variable force F is: (A) mgL(1-sin) (B) mgL (C) mgL(1-cos) (D) mgL(1+cos) Solution : Work done by the external force =  E of the system (object) = PE ( ) = mgh = mg (L – L cos ) = mgL (1 – cos ). (C) 10: When a body of mass M slides down an inclined plane of inclination , having coefficient of friction  through a distance s, the work done against friction is: (A)  Mg cos  s (B)  Mg sin  s (C) Mg ( cos   sin )s (D) None of the above Solution : (A)Work done against friction = - work done by friction = mg cos  s. 11: 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 frictional force will be (A) push effect. (B) pull effect (C) remain same. (D) data insufficient. Solution : (D)fr = (mg  F sin) pulling. Fv = (mg + F sin) pushing. 12: A body is acted upon by a force which is inversely proportional to the distance covered. The work done will be proportional to: (A) s (B) s2 (C) s (D) None of the above Solution : (D)W =  F.ds= = kln (s/s1) 13: 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  = 600 with horizontal. If the particles move together, the velocity of the combination just after the impact is equal to (A) v (B) v/2 (C) (D) v/4 Solution : (D)mv cos  = (M + m) v  v = cos 600 = . 14: 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 (B) 5 m/s (C) 3.13 m/s (D) 3.6 m/s Solution : (C)mg = mv2 =  v =  3.13 m/s. 15. A neutron mass 1.67  10–27 kg moving with a velocity 1.2  107 ms–1 collides head-on with a deuteron of mass 3.34  10–27 kg initially at rest. If the collision is perfectly inelastic, the speed of the composite particle will be (A) 2  106 ms–1 (B) 4  106 ms–1 (C) 6  106 ms–1 (D) 8  106 ms–1 Sol: (B) Mass of neutron (m) = 1.67  10–27 kg. Speed of neutron (v) = 1.2  107 ms–1. Notice that the mass of deuteron (M) = 3.34  10–27 kg = 2m. If V is the speed of the composite particle, the law of conservation of momentum gives or or . 16. In above. Q. if the collision were perfectly elastic, what would be the speed of deuteron after the collision ? (A) 2  106 ms–1 (B) 4  106 ms–1 (C) 6  106 ms–1 (D) 8  106 ms–1 Sol: (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 . 17. 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 (B) 6000 J (C) 6249 J (D) 6145 N Solution: (A)  Here we will calculate acceleration a = = = 8 m/s2 Velocity at the end of 5 seconds is v = u + at v = 0 + 85 = 40 m/s  K.E. acquired = =  5 (40)2 = 4000J 18. A body is constrained to move along z-axis of the co-ordinate system is being applied by a constant force . Find the work done by this force in moving the body over a distance of 5m along z-axis. (A) 30 J (B) 20 J (C) 40 J (D) 61 J Solution: (B) Here displacement is only along z-axis Hence, and  Work done by force W = = . = 20J 19. 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 done by the force (A) 50 J (B) 60 J (C) 40 J (D) 45 J Solution: (D) Work done dw =  W = = Area under F-S curve =  5 10 + 210 = 45J 20. A body dropped forma height ‘H’ reaches the ground with a speed of 1.1 . Calculate the work done by air friction? (A) 0.395 mgH (B) 0.395 mgH (C) 0.400 mgH (D) 0.400 mgH Solution : (D) K.E. of particle on reaching the ground = m.(1.21)gh = 0.605mgH Now KE = mgH  0.605 mgH = mgH   = 0.395 mgH Now since body moves opposite to the direction of force of air friction.  (air friction) = 0.395 mgH 21. A block of mass m = 2kg is attached to two unscratched springs of force constant k1 = 100 N/m and k2 = 125 N/m. 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 (B) 1.02 m/s (C) 1.04 m/s (D) 1.05 m/s Solution: (A) Here effective spring constant of the combination = k1 + k2 = 225 N/m Now applying the energy conservation principle, we have [100+125]  (0.1)2 = mv2 [100+125]  (0.1)2 =  2  v2  v2 = 1.125  v = 1.06 m/s at the mean position. 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 100N/m, then how much is the spring compressed? (A) 1. 7 m (B) 1. 6 m (C) 1. 4 m (D) 1. 8 m Solution: (B) Here energy is conserved.   X = = = 1.6 m 23. A small block of mass 500 gm is pressed against a horizontal spring fixed at one end to a distance of 10 cm. when released, the block moves horizontally till it leaves the spring. Where it will hit the ground 5 m below the spring. (Take k = 100 N/m of spring) (A) 1.441 m (B) 1. 6 m (C) 1.414 m (D) 1. 10 m Solution: (C) v horizontal of block = vH = = m/s Now gt2 = 5 mt = 10/10 = 1sec (where g = 10 m/s2) so RH = vH  1 = 1 = 1.414 m Block will hit the ground at 1.414 m from a point 5m below the spring. 24. A block of mass 0.5 kg is kept on a rough inclined plane making an angle of 30 with horizontal. What power will be required to move the block up the plane (along the plane) with a velocity of 5m/s? (Take  = 0.2 between block and plane) (A) 16.825 N-m/s (B) 16.822 N-m/s (C) 16.852 N-m/s (D) 16.528 N-m/s Solution : (A)The total downward force acting on the block = 0.510 = 5[0.5 + 0.173] = 3.365 N. Now the power required to move up along the inclined power = 3.365  5 = 16.825 N-m/s 25. A small body of mass m is located on a horizontal plane. The body acquires a horizontal velocity v0 . Find mean power developed by the frictional force, during the whole time of its motion. If coefficient friction x = 0.27, mass of body m = 1kg, and V0 = 1.5 m/s (A) (B) (C) (D) Solution: (B) Work done against friction w = - mgs P = = – mgs. = – mgs.v Pav = = 26. A block of mass m moving with a velocity v0 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 (A) (B) (C) (D) Solution : (A)Applying work energy theorem PE+ KE = 0  kx2 - m =0  x =  Fmax = kx = 27. 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. (A) a = (B) a = (C) a = (D) a = Ans. (B) Solution : Speed will be maximum, when engine of bus will produce maximum power i.e. P  P = R.vmax  vmax = Here v = Force by which engine pulls the bus is P = F.  F = = 2 R  Since F – R = m.a  2 R – R = m.a a = 28. A uniform rod of length L and mass m hinged at one end is hanging vertically. The other end is now raised until it makes an angle 60o with vertical. How much work is required? (A) (B) (C) (D) Ans. (D) Solution: We can consider its whole mass is centre ted at its centre of mass O.  work done = mgh = mg = mg = 29- 32. 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 29. The change in potential energy of the mass with angle (A) U =  mgR(1  cos) (B) U = + mgR(1  cos) (C) U = + mgR(1 + cos) (D) U =  mgR(1 + cos) Ans. (A) Solution: Consider the mass when it is at the point B. UA (P.E. at A) = 0 UB(P.E. at B) = mgR (1  cos)  U = UB  UA  U =  mgR(1  cos) Negative sign indicates that P.E. decreases as particle slides down. 30. The kinetic energy as a function of angle (A) T = mgR(1  cos) (B) T = mgR(1 + cos) (C) T = – mgR(1  cos) (D) T = gR(1  cos) Ans. (A) Solution: Conserving energy at points A and B. UA + TA = UB + TB where UA = P.E. at A, UB = P.E. at B TA = K.E. at A, TB = K.E. at B  0 + 0 = mgR(1 cos) + TB  T = mgR(1  cos) 31. The radial and tangential acceleration as a function of angle. (A) at = sin (B) at = gsin (C) at = g (D) at = –gsin Ans. (B) Solution: Since T =  = mgR(1 - cos2mgR sin2  v = sin/2  aradial = v2/R  ar = 4g sin2/2 at = g sin also, in circular motion velocity is along the tangent, therefore vtangential = 2 sin/2  at =  at = (gR) cos/2 = (gR) cos/2  at = (g/R) v cos/2 , as R = v  at = gsin 32. The angle at which the mass flies off the sphere. (A) = cos-1(4/3) (B) = cos-1(1/3) (C) = cos-1(2/3) (D) = cos-1(2/8) Ans. (C) Solution : For circular motion mg cos - N = at the moment when the particle brakes off the sphere N = 0.  mg cos =  g cos = v2/R  v = 2 sin/2 g cos = 4g sin2/2 = 2g(1  cos)  cos = 2/3  = cos-1(2/3) 33-37. A block of ice mass 10 kg slides down an incline 5 m long and 3 m high. 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 : 33. The work done by the man on the block (A) Wm = –300 J (B) Wm = –260 J (C) Wm = –400 J (D) Wm = –560 J Ans. (B) Solution : From the figure sin = 3/5 and cos = 4/5 F.B.D. of the block F = force by the man  = frictional force N = normal reaction of the surface mg = gravity Since block slides with constant speed, hence mg sin= F +   F = mg sin -  = 10  10  , as f =  mg cos (A) Wman = Here F = 52 N, and S = 5m.  Wm = -52  5 J = -260 J 34. The work done by gravity on the block (A) Wgravity = 300 J (B) Wgravity = 400 J (C)Wgravity = 600 J (D) Wgravity = 500 J Ans. (A) Solution : Wgravity = mg S sin= 10  10 5  (3/5) J = 300 J 35. The work done by the surface on the block (A) Wsurface = + 40 J (B) Wsurface = - 42 J (C) Wsurface = - 41 J (D) Wsurface = – 40 J Ans. (D) Solution : Wsurface = WN + Wfriction = 0 + fS cos180o = 0 - mg cos S = -0.1  10  10  (4/5)  5 J = -40 J 36. The work done by the resultant forces on the block (A) W = 0 (B) W = 2 (C) W = 3 (D) W = 6 Ans. (A) Solution : Work done by the resultant force is given by W = Wm + Wg + WN + Wf = -260 J + 300 J + 0 - 40 J = 0 37. The change in K.E. of the block . (A) K.E. = 0 (B) K.E. = 6 (C) K.E. = 4 (D) K.E. = 3 Ans. (A) Solution : Since net work done by all the force is zero, hence change in K.E. = 0. 38: If a constant force is acting on a particle and displacement of particle becomes . Find net work done by force . (A) 55 J (B) 80 J (C) 90 J (D) 100 J Solution: (A) 39: 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) 100J (B) 200J (C) 300J (D) 400J Work is zero because displacement is zero. Correct solution: (A) Force F is varying because its direction is changing so where dr is a small path length and small path length will always be along tangent 40: 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 l0 = 1 mt. and x = 25mt initially. (A) (B) (C) (D) Solution: In this case force F is variable Common mistake: = [F cos i + F sin j]. dx i = F cos dx cos = We know that in this case angle between and displacement is acute so work should be positive, contradiction has come. Correct solution: (B) y = 25 – x dy = – dx and dw = (F cos i + F sin j). dy i dw = –Fcos dx 41: A projectile is projected with initial velocity 20m/s and at an angle of 30 from horizontal. Find the total work done when it will hit the ground. (A) 0 (B) 10 (C) 20 (D) 30 Solution: (A)work = (mg j) . (r i) = 0 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 = 3t2. Find work done by force F when it will move 2mt. (A) 12 J (B) 16 J (C) 32 J (D) 42 J Solution: (A) x = 3t2 Force acting on particle = ma = 1  6 = 6N Displacement = 2 work = 43: If a projectile is projected with initial speed u and angle from horizontal then what will be its average power up to time when it will hit the ground again. (A) 100 (B) 200 (C) 300 (D) 0 Solution:(D) When it will hit the ground the net work done is zero. So 44: 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. (A) (B) (C) (D) Solution: (C) Fv = constant = c 45. A fast moving neutron suffers one-dimensional elastic collision with a nucleus . Wha approximate percentage of energy is lost by the neutron in the collision ? (A) 5 % (B) 10 % (C) 25 % (D) 0 % Solution: (C) 46. An 8 kg block accelerates uniformly from rest to a velocity of 4 ms–1 in 40 second. The instantaneous power at the end of 8 second is (A) 0.64 W (B) 0.32 W (C) 0.16 W (D) 0.08 W. Solution: (A) Velocity at the end of 8 second Required power = (ma) 47: Work is which type of physical quantity. (A) vector (B) scalar (C) tenser (D) None of these. Solution : (B) 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. (A) (B) (C) (D) Solution: (A) At the time of maximum velocity f = r. i.e., 49. Block m has given a velocity v0 towardly right. What will be the maximum power delivered by the spring force during the entire motion. (A) (B) (C) (D) Solution: In the given case v = v0 cos(t) ……(i)  x = …….(ii) Acceleration = - v0  sin(t) …….(iii) Force = mass  acceleration = mv0 sin(t) Power = - force v = mv02 sin(t) cos (t) P =  max power P max = for t =  minimum power So Ans = (C) 50: A particle of mass 10kg is moving with velocity 2i + 3j + 4k what will be its kinetic energy. If A constant force starts acting on body then find after 2 second its power and work done by this force in 2 second. (A) (B) (C) (D) Solution: Acceleration = Work = = (D) 51. 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) (B) (C) (D) Solution: So work by spring = (B) 52: Check whether the force is conservative or not. (A) conservative (B) non conservative (C) variable (D) none of these. Solution: Let a particle on which force is acting moving from (x1, y1, t1) to (x2, y2, t2) then dw = .d and d = dxi + dyj + dzk dw = (xi + yj = zk).(dxi + dyj + dzk) So above work depend only on the coordinates of point (x1, y1, z1) and (x2, y2, z2). So force is conservative. (A) 53: Prove that the central force which follows the inverse square law are conservative forces. (A) (B) (C) (D) Solution: Let consider a central force . Now we will prove that if a body on which the force is acting , will move from A to B then work done on it will not depend on both , (B) 54: Prove that is a conservative force. (A) w = (B) w = (C) w = (D) w = Solution: Let particle moves from (x1, y1, z1) to (x2, y2, z2) dw = x2 dx + y2dy + z2dz w = which depend only on (x1, y1, z1) and (x2, y2, z2) So will be conservative. (A) 55: 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 ac= k2r t2 where k is constant what is the power delivered to the particle by the forces acting on it. (A) (B) (C) (D) Solution Centripetal force will not supply the power and power by the tangential force = m (P = FV) = mk2rt2 = (C) 56: Find the total compression in the spring. (A) (B) (C) (D) Solution: E Ti = ETf – work done by nonconservative force E Ti = E Tf = w = - (A) 57. 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. between time t = 2 and t = 4 sec. (A) 212 J (B) 213 J (C) 214 J (D) 215 J Solution: (C) P = maV = mv = 4t3 – 5t + 2 K. E. = = t4 – + 2t So change in K.E. will be 214 J. (C) 58: 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) 4s (B) 5s (C) 8s (D) 10s Solution: (C) (C) 59: What is work done when body will displace perpendicular to the force. (A) zero (B) maximum (C) minimum (D) None of these. Solution: (A) W = F.d = Fd cos  = 90o then W = 0 (A) 60: 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 Solution: (B) Mass of water pumped per second Kinetic energy of water coming out per second  Power of pump = 180 J /s = 180 W (B)