Title 07. Work, Energy and Power Med Ans.pdf ✅

1. (d) : Here, m =1 3 g 10− = kg H=1km = 1000 m = 10 m 3 The change in kinetic energy of the drop is mv 0 2 1 K 2  = − (u = 0) 10 50 50 1.25J 2 1 3 =    = − The work done by the gravitational force is W mgh 10 10 10 10J 3 3 g = =   − According to work-energy theorem K = wg + Wr Where Wr is the work done by the resistive force on the raindrop. Thus W K W 1.25J 10J 8.75J r =  − g = − = − 2. (a) : Let v is the speed of the object at the bottom of the plane. According to work-energy theorem w = K = Kf − Ki 2 2 mu 2 1 mv 2 1 mgh = − (u = 0) 2 mv 2 1 mgh = (u = 0) or v = 2gh From the above expression, it is clear that v is independent of the mass of an object. 3. (b) : Work done, W = F.d = Fdcos   Where  is the angle between the force F  and displacement d  (i) As the body moves along the direction of applied force. 0  = 0 , W = Fdcos 0 0 =Fd. It is positive, (ii) As the body moves in a direction opposite to the gravitational force which acts vertically downwards. 180 , 0  = W = Fdcos 0 180 = - Fd. It is negative. 4. (b) : Here, k unit ˆ j 5 ˆ i 4 ˆ F = 3 + −  k ˆ j 3 ˆ i 4 ˆ d = 5 + +  unit | F| (3) (4) ( 5) 50 unit 2 2 2  = + + − =  | d | (5) (4) (3) 50 unit 2 2 2 = + + =  Let  be angle between Fand d   . 50 50 k) ˆ j 3 ˆ i 4 ˆ j 5).(5 ˆ i 4 ˆ (3 | F || d | F.d cos + − + +   =   =   or cos (0.32) −1  = 5. (b) : Here, As the body moves a distance of 4 m along y-axis, Work done by the given force is 6. (a) : work against friction is = 8.7J 7. (c) : Work done by applied force is W = Fd = 10N 10m = 100 J 8. (d) : 1 J 10 erg 7 = 9. (c) : When a weightlifter lifts a weight work done by the lifting force, W Fdcos0 Fd 0 1 = = Work done in holding it up W2 = 0 (because displacement is zero) 10. (c) : Mass of length 2 m of the chain = 4kg Mass of length 60 cm or 0.60 m of the chain 1.2kg 2 4 0.60 =  = Weight of the hanging part of the chain = 1.2 × 10 = 12 N Since the centre of gravity of the hanging part lies at its mid point, i.e. 30 cm or 0.30m W =12  0.30 = 3.6J 11. (d) : The weight of hanging part       3 L of chain is       mg 3 1 . This weight acts at centre of gravity of the hanging part, which is at a distance of       6 L from the table. As work done = force × distance 18 MgL 6 L 3 Mg W =  = 12. (c) : kinetic energy, 2m p k 2 = Since m remains constant 2 K  p 2 1 2 1 1 2 2 1 2 1 or k 4k 4 1 2p p p p k k = =         =         = Percentage increase in kinetic energy 100 300% k 4k k 100 k k k 1 1 1 1 2 1  = −  = − = 13. (a) : Volume of wind flowing per second = Av Mass of wind flowing per second = Av F = − ˆ i + 2 ˆ j+3k ˆ N  d ˆ = 4 ˆ jm jm) 8J ˆ kN).(4 ˆ j 3 ˆ i 2 ˆ W = F.d = (− + + =   W mgd sin (1kg) (10ms ) sin 30 10m 50J 2 0 g =  =    = − Wf = fd = Nd = mg cos d (N = mg cos) 0.1 1kg 10ms cos30 10m 2 0 =     −
Mass of air passing in time t s = Avt 14. (c) : Kinetic energy of the air 2 2 (Av t)v 2 1 mv 2 1 = =  Av t 2 1 3 =  15. (a) : If an object of mass m moving with velocity v  , its kinetic energy k is given by 2 mv 2 1 mv.v 2 1 k = =   As mass m and v (v.v) 2   are always positive, therefore kinetic energy is always positive. The kinetic energy can never by negative. 16. (d): Here, m 120g 10 kg 0.12kg 3 =  = − 1 jms ˆ i 5 ˆ v 2 − = +  2 2 2 j) 4 25 29m s ˆ i 5 ˆ j).(2 ˆ i 5 ˆ v v.v (2 −  = = + + = + =   Kinetic energy, 2 m.v 2 1 k = 0.12kg (29 m s ) 2 1 2 −2 =   =1.74J 17. (a) : Friction always opposes motion. A body does work against friction at the expense of its kinetic energy. Hence, the work done by a body against friction always results in loss of kinetic energy. 18. (d) : The average energy consumed by a human being in a day is 2400 kcal 4.2 10 J/ kcal 10 J 3 7 =  = 19. (c): work done = Force × distance = Change in kinetic energy. Both the truck and the car had same kinetic energy and hence same amount of work is needed to be done. As retarding force applied is same for both, therefore, both the truck and the car travel the same distance before coming to rest 20. (b): Here, m1 = 20kg,m2 = 5kg Force = 40 N 2 2 2 a1 2ms and a 8ms − −  = = 2 2 2 B 2 1 1 A m (a t ) 2 1 m (a t ) 2 1 = 2 B 2 A 5 64 t 2 1 20 4 t 2 1     =    1 4 20 4 2 1 5 64 2 1 t t 2 B 2 A =      = 2 t t B A = 21. (a) : v 1 F  v C f = Where C is a constant of proportionality. v C dt dv or m v C  ma = = m Cdt vdv = Integrating both sides, we get mv Ct 2 1 or m Ct 2 v 2 2 = = Or Kinetic energy, mv Ct 2 1 k 2 = = Or K  t 22. (d) : The area under force – displacement curve represents work done. 23. (c) : Potential energy of the spring when stretched through a distance x, kx 10J 2 1 U 2 = = When x becomes 2x, the potential energy will be kx 4 10 40J 2 1 k(2x) 4 2 1 U' 2 2 = =  =  =  work done = U'−U = 40 −10 = 30J 24. (d) : t wh t mgh time enegy time work Power = = = = (w = weight) 9 11 12 11 3 4 t t W W P P 1 2 2 1 2 1  =  =  = (h is same in both cases) 25. (b) By the system against a conservative force 26. (a) : Let the ball is dropped from height h, Initial potential energy = mgh. Potential energy after first bounce = mg × 80% of h = 0.80 mgh. Potential energy lost in each bounce = 0.20 mgh Fraction of potential energy lost in each bounce 0.20 mgh 0.20mgh = = 27. (c) : Potential energy 28. (a) : Among the given forces, force of friction is a non- conservative force whereas all other forces are conservative forces. 29. (c) : Friction is a non-conservative force. Work done by a non- conservative force over a closed path is not zero.
Hence, option (c) is a false statement. While all other statements are correct. 30. (b) : Since the particle is moving in horizontal circle, centripetal force, 2 2 r k r mv F = = r k mv 2 .....(i) Kinetic energy of the particle, 2r k mv 2 1 k 2 = = (using (i)) dr du As F − =  potential energy, r k dr k r dr r k U Fdr r 2 r 2 r −  = =      − = = −     −    total energy = 2r k r k 2r k K U − + = − = 31. (b) : The situation is as shown in the figure. Potential energy of the bob at A = mgh Kinetic energy of the bob at 2 mv 2 1 B = As 10% of the energy of the bob is dissipated against air resistance. 100 90 mv 2 1 2  = of mgh or g h 100 90 v 2 2 =    1 6 ms 100 2 90 10 2 100 2 90 g h v − =    =    = 32. (c) : Total mechanical energy at height, H EH = mgh Let h v be velocity of the ball at height h       = H 4 3  total mechanical energy at height h, 2 h mv h 2 E mgh  = + According to law of conservation of mechanical energy. EH = Eh 2 mv h 2 1 mgH = mgh + v 2g(H h) 2 h = − Required ratio of kinetic energy to potential energy at height h is mgh m2g(H h) 2 1 mgh mv 2 1 V K 2 h h h − = =       = − − = 1 h H h H h       = −1 3 4 H) 4 3 (h = 3 1 = 33. (d) : All the central forces are conservative. All other statements are correct. 34. (b) : When 1 kg of coal is burnt the energy released is 30 MJ (=3×107 J). 35. (d) : In exothermic reaction heat is released. In endothermic reaction heat is absorbed. Energy released in burning 1 litre of gasoline is30 MJ ( 3 10 J) 7 =  36. (d) : The energy required to break one bond in DNA is 10 J −20 37. (b) : Among the given units watt is the unit of power whereas all others are the units of energy. 38. (a) : Here, m = 60 + 20 = 80 kg h = 20 × 0.2 = 4 m g = 9.8ms ,t 10s 2 = − 313.6W. 10 3136 10 80 9.8 4 t mgh t W p = =   = = = 39. (c) : 1 kilowatt hour (kW h) = (103 W) × (3600 s) eV 1.6 10 3.6 10 19 6 −   = ( 1eV 1.6 10 J) −19  =  = 2.25 10 eV 25  40. (a) : Force, F = (4 ˆ i + ˆ j− 2k ˆ )N  Velocity, 1 k)mS ˆ j 3 ˆ i 2 ˆ v (2 − = + +  Power, k) ˆ j 3 ˆ i 2 ˆ k).(2 ˆ i j 2 ˆ P = F.v = (4 + − + +    = (8 + 2 − 6)W = 4W 41. (b) : Here, m = 1800 kg Frictional force, f = 4000 N Uniform speed, v = 2m s-1 Downward force on elevator is F = mg+f = (1800kg 10ms ) 4000N 22000N 2  + = −
The motor must supply enough power to balance this force. Hence, p Fv (22000N)(2ms ) −1 = = = 44000 W = 44 ×103 W = 44 kW 42. (b) : Using v = u + at v = at (u = 0) As power, p = F × v p (ma) at ma t 2  =  = As m and a are constants p  t 43. (b) : Here, Volume of water 3 = 30m t = 15min = 15  60s = 900s Height, h = 40m Efficiency,  = 30% Density of water = 3 3 10 kg m − Mass of water pumped = Volume × Density (30m )(10 kgm ) 3 10 kg 3 3 3 4 = =  − 900s (3 10 kg)(10ms )(40m) t mgh t W P 4 2 output −  = = = 10 W 3 4 4 =  Efficiency, input output P P  = ifj.kke ifj.kke P P  = 5 4 Output input 10 9 4 100 30 3 P 4 10 P =   =  = = 44.4103W = 44.4KW 44. (d) : t mgh P = or g h p t m   = 1200 kg 1200 litre 10 ms 10m 2000 W 60s m 2 = =   = − 45. (b) : Power = kinetic energy of water flowing per second 2 1 = ×area of cross-section × density × (velocity)3 50 5 1000 2 W 10 W 1MW 2 1 3 6 =     = = 46. (a) : According to momentum conservation, we get m1v1i = m1v1f + m2v2f ......(i) Where 1i v is the initial velocity of spherical ball of mass m1 before collision and 1f v2f v and are the final velocities of the balls of masses m1 and m2 after collision. According to kinetic energy conservation, we get m v f 2 1 m v 2 1 m v 2 1 2 2 2 2 1 1 2 1 1i = + 2 2 2f 2 1 1f 2 m1v1i = m v + m v .....(ii) Form Egs. (i) and (ii), it follows that m v (v v ) m v (v v ) 1 1i 2f − 1i = 1 1 2f − 1f Or v (v v ) v i v (v v )(v v ) 1i 1f 1i 1f 2 1f 2 2f 1i − 1f = 1 − = − + 2f 1i 1f v = v + v Substituting this in Eq. (i) we get 1i 1 2 1 2 1f v m m (m m ) v + − = The initial kinetic energy of the mass m1 is 2 1i 1 1i m v 2 1 k = The final kinetic energy of the mass m1 is 2 1i 2 1 2 1 2 1 2 1f 1 1f v m m m m m 2 1 m v 2 1 k         + − = = (Using (iii)) The fraction of kinetic energy lost by m1 is 1i 1i 1f k k k f − = 2 1 1i 2 1i 2 1 2 1 2 1 2 1 1i m v 2 1 v m m m m m 2 1 m v 2 1         + − − = 2 1 2 1 2 2 1 2 1 2 (m m ) 4m m m m m m 1 + =         + − = − 47. (c) : for A: Using the laws of conservation of linear momentum and energy the velocities of two bodies after perfectly elastic collision are given by 1 2 2 2 1 2 1 2 1 m m 2m u m m (m m ) v + + + − = .......(i) 1 2 2 1 2 1 2 1 1 2 m m (m m )u m m 2m u v + − + + = .......(ii) Where 1 m2 m and are the masses of two bodies and 1 u2 u and are the velocities of two bodies before collision. If m1 = m2 then Form (i) and (ii), we get v1 = u2 and v2 = u1 i.e., in perfectly elastic collision of two moving bodies of equal masses, the bodies merely exchange their velocities after collision. Thus, statement A is correct. For B. In Eqs. (i) and (ii) If m2  m1 andu2 = 0 then 1 1 v2 2u1 v = u and = Thus statement B is wrong. 48. (d) : When two spheres of equal masses undergo a glancing elastic collision with one of them at rest, after the collision they will move at right angle to each other.