Tiêu đề 6.WORK ENERGY AND POWER - Explanations.pdf ✅

1 (a) Work done = area under curve and displacement axis = 1 × 10 − 1 × 10 + 1 × 10 = 10 J 2 (c) Force produced by the engine F = P v = 30×103 30 = 103N Acceleration = Forward force by engine − resistive force mas of car = 1000 − 750 1250 = 250 1250 = 1 5 m/s 2 3 (a) 1 2 kS 2 = 10 J [Given in the problem] 1 2 k[(2S) 2 − (S) 2 ] = 3 × 1 2 kS 2 = 3 × 10 = 30 J 4 (b) Minimum force mg sin θ, so, minimum power is given by P = mg sin θ v or v = P mg sin θ or v = 9000×2 1200×10×1 ms −1 = 15ms −1 = 15 × 18 5 = 54 kmh −1 5 (b) Loss of KE = force × distance = (ma)x As a ∝ x ∴ Loss of KE ∝ x 2 6 (b) Power ,p= Total Energy t = mgh+ 1 2 mv2 t 10 × 10 × 20 + 1 2 + 10 × 10 × 10 1 =2000+500=2500 W=2.5 KW 7 (b) In case of elastic collision ,coefficient of restitution e=1 or Relative speed of approach =relative speed of separation. ∴ Option (b)is correct. 8 (d) Initial momentum = P⃗ = mvî+ mvĵ |P⃗ | = √2mv Final momentum = 2m × V By the law of conservation of momentum 2m × V = √2 mv ⇒ V = v √2 In the problem v = 10 m/s [Given] ∴ V = 10 √2 = 5√2 m/s 9 (a) The ball rebounds with the same speed. So change in it’s Kinetic energy will be zero i. e. work done by the ball on the wall is zero 10 (a) If a body has momentum, it must have kinetic energy also, (a) is the wrong statement If the energy is totally potential, it need not have momentum (b) is correct (c) and (d) are also correct 11 (b) Initial velocity of particle, vi = 20 ms −1 Final velocity of the particle, vf = 0 According to work-energy theorem, Wnet = ∆KE = Kf − Ki = 1 2 m(vf 2 − vi 2 ) = 1 2 × 2(0 2 − 202 ) = −400 J 13 (b) If the masses are equal and target is at rest and after collision both masses moves in different direction. Then angle between direction of velocity will be 90°, if collision is elastic 15 (d) Friction is a non-conservative force 16 (a) Given,r1 = 2î− 3ĵ− 4k̂ And r2 = 3î− 4ĵ+ 5k̂ Now, r2 − r1 = î− ĵ+ 9k̂ And F = 4î+ ĵ+ 6k̂ ∴ work done = F.r W = (4î+ ĵ+ 6k̂). (î− ĵ+ 9k̂) = 4 − 1 + 54 = 57 J 17 (a) ( ∆k k ) retained = ( m1 − m2 m1 + m2 ) 2 = ( 1 − A 1 + A ) 2 18 (d) Mass to be lifted = 10 × 102kg [∴ density of water = 103kgm−3 ] Height, h = 10 m Work done = 104 × 10 × 10 = 106 J 19 (b) As the area above the time axis is numerically equal to area below the time axis therefore net momentum gained by body will be zero because momentum is a vector quantity 20 (d) Let m be the mass of the block, h the height from which it is dropped, and x the compression o the spring. Since, energy is conserved, so Final gravitational potential energy Neutron Nucleus at rest m1 = 1 m2 = A
= final spring potential energy or mg(h + x) = 1 2 kx 2 or mg(h + x) + 1 2 kx 2 = 0 or kx 2 − 2mg(h + x) = 0 kx 2 − 2mgx − 2mgh = 0 This is a quadratic equation for x. Its solution is x = mg ± √(mg) 2 + 2mghk k Now, mg = 2 × 9.8 = 19.6 N and hk = 0.40 × 1960 = 784 N ∴ x = 19.6 ± √(19.6) 2 + 2(19.6)(784) 1960 = 0.10 m or −0.080 m Since, x must be positive (a compression) we accept the positive solution and reject the negative solution. Hence, x = 0.10 m 21 (b) In case of non-conservative forces, the work done is dissipated as heat, sound etc. ie, it does not increase the potential energy. But in case of conservative forces, work done is responsible for increasing the potential energy 22 (c) Let the thickness of one plank be s If bullet enters with velocity u then it leaves with velocity v = (u − u 20 ) = 19 20 u From v 2 = u 2 − 2as ⇒ ( 19 20 u) 2 = u 2 − 2as ⇒ 400 39 = u 2 2as Now if the n planks are arranged just to stop the bullet then again from v 2 = u 2 − 2as 0 = u 2 − 2ans ⇒ n = u 2 2as = 400 39 ⇒ n = 10.25 As the planks are more than 10 so we can consider n = 11 23 (b) W = mg sin θ × s = 2 × 103 × sin 15° × 10 = 5.17 kJ 24 (a) K.E. acquired by the body = work done on the body K. E. = 1 2 mv 2 = Fs i. e. it does not depend upon the mass of the body although velocity depends upon the mass v 2 ∝ 1 m [If F and s are constant] 25 (b) Let the initial mass of body = m Initial linear momentum = mv ...(i) When it breaks into equal masses then one of the fragment retrace back with same velocity ∴ Final linear momentum = m 2 (−v) + m 2 (v2) ...(ii) By the conservation of linear momentum ⇒ mv = −mv 2 + mv2 2 ⇒ v2 = 3v i. e., other fragment moves with velocity 3v in forward direction 26 (d) Let a nucleus of mass M splits into two nuclear parts having masses M1 and M2 and radii R1 and R2 and densities ρ1 and ρ2 ∴ M1 = ρ1 4 3 πR1 3 and M2 = ρ2 4 3 πR2 3 Given: ρ1 = ρ2 ∴ M1 M2 = ( R1 R2 ) 3 According to law conservation of linear momentum, M × 0 = M1v1 + M2v2 or M1 M2 = − v2 v1 −ve sign show that both the parts are move in opposite direction in order to conserve the linear momentum ∴ v1 v2 = M2 M1 or v1 v2 = ( R2 R1 ) 3 v1 v2 = ( 2 1 ) 3 = 8 1 [Given R1 R2 = 1 2 ] 28 (d) Conservation of linear momentum along x- direction m/2 v1 = -v m/2 v2 After explosion m v Before explosion
m2v = m1vx m2v m1 = vx Along y direction m2 × v 2 = m1vy tan θ = 1 2 29 (a) Initially U 238 nucleus was at rest and after decay its part moves in opposite direction According to conservation of momentum 4v + 234V = 238 × 0 ⇒ V = − 4v 234 30 (c) Work done W = mgh + ∆pV =Vpgh + ∆pV Given V=4m3 , p = 103kg −2 , g = 10ms −2 , h = 20m, ∆p = (2 × 105 − 1 × 105)Nm−2 W = 4 × 103 × 10 × 20 + (2 × 105 − 1 × 105 ) × 4 8 × 105 + 4 × 105 = 12 × 105 J 31 (d) Here, m1 = 20 kg, m2 = 0.1 kg, v1 = velocity of recoil of gun, v2 = velocity of bullet As m1v1 = m2v2 v1 = m2 m1 v2 = 0.1 20 v2 = v2 200 Recoil energy of gun= 1 2 m1v1 2 = 1 2 × 20 ( v2 200 ) 2 804 = 10v2 2 4 × 104 = v2 2 4 × 103 v2 = √804 × 4 × 103ms −1 32 (a) U1 = mgh1 and U2 = mgh2 % energy lost = U1−U2 U1 × 100 = mgh1 − mgh2 mgh1 100 = ( h1 − h2 h1 ) × 100 = 2 − 1.5 2 × 100 = 25% 33 (a) Velocity of 50 kg mass after 5 sec of projection v = u − gt = 100 − 9.8 × 5 = 51 m/s At this instant momentum of body is an upward direction Pinitial = 50 × 51 = 2550 kg − m/s After breaking 20 kg piece travels upwards with 150 m/s let the speed of 30 kg mass is V Pfinal = 20 × 150 + 30 × V By the law of conservation of momentum Pinitial = Pfinal ⇒ 2550 = 20 × 150 + 30 × V ⇒ V = −15 m/s i. e. it moves in downward direction 34 (c) v = √(8) 2 + (6) 2 = 10ms −1 KE = 1 2 mv 2 = 1 2 × 0.4 × 10 × 10 = 20 J 35 (b) Ek ′ Ek = (1 + 1 5 ) 2 = 36 25 ( Ek ′ Ek − 1) × 100 = ( 36 26) × 100 = 44 36 (d) Velocity at B when dropped from A where AC = S v 2 = 0 + 2g(S − x) Or v 2 = 2g(S − x) ...(i) Potential energy at B = mgx ...(ii) ∵ Kinetic energy = 3 × potential energy ∴ 1 2 m × 2g(S − x) = 3 × mgx ⇒ S − x = 3x or S = 4x or x = S/4 From (i), v 2 = 2g(S − x) = 2g (S − S 4 ) = 2g × 3S 4 = 3gS 2 ⇒ v = √ 3gS 2 ∴ x = S 4 and v = √ 3gS 2 37 (d) Given F = 2x, Work done W = ∫ F dx ∴ W = ∫ 2x dx = 2 [ x 2 2 ] x2 x1 x1 x2 =(x2 2 − x1 2 ) 38 (d) Watt and Horsepower are the units of power 39 (a) When two bodies of same mass makes head on elastic collision, and then they interchange their velocities. So, after collision first body starts to move with velocity v. 40 (c) According to law of conservation of momentum 234 V  particle Residual nucleus v 4
Momentum of neutron = Momentum of combination ⇒ 1.67 × 10−27 × 108 = (1.67 × 10−27 + 3.34 × 10−27)v ∴ v = 3.33 × 107 m/s 41 (c) Power = W t If W is constant then P ∝ 1 t i. e. P1 P2 = t2 t1 = 20 10 = 2 1 42 (b) From conservation of energy, Potential energy at height h = kinetic energy at ground Therefore, at height h, potential energy of ball A PE = mAgh KE at ground = 1 2 mAvA 2 So, mAgh = 1 2 mAvA 2 vA = √2gh Similarly, vB = √2gh Therefore, vA = vB 43 (d) As the speed of mass is uniform hence, net power will be zero. 44 (b) Potential energy of water = kinetic energy at turbine mgh = 1 2 mv 2 ⇒ v = √2gh = √2 × 9.8 × 19.6 = 19.6 m/s 45 (d) At a given height the half of the kinetic energy of the body is equal to its potential energy. Initial kinetic energy of the body = 1 2 mv 2 = 1 2 m(4) 2 = 8m Let at height h, the kinetic energy reduces to half, i. e., it becomes 4 m. It is also equal to potential energy. Hence, mgh = 4m or h = 4 g = 4 10 = 0.4 m 46 (a) Since body moves with constant velocity, so. Net force on the body is zero. Here, N = mg,F = f ∴ W = F . s = fs cos 180′′ = fs = −10 × 2 = −20 J 47 (d) In compression or extension of a spring work is done against restoring force In moving a body against gravity work is done against gravitational force of attraction It means in all three cases potential energy of the system increases But when the bubble rises in the direction of upthrust force then system works so the potential energy of the system decreases 48 (c) P = Fv = m. dv dt ∙ v ∫ v dv = ∫ p mdt; v 2 2 = pt m v = √ 2p m t 1/2 ; dx dt = √ 2p m t 1/2 ∫ dx = √ 2p m ∫ t 1/2dt; x = √ 2p 3 t 3/2 3/2 = 2 3 √ 2p 3 t 3/2 x ∝ t 3/2 49 (d) From v = u + at, v1 = 0 + at1 (∵ a = v1 t1 ) F = ma = mv1/t1 Velocity acquired in t sec = at = v1 t1 t Power = F × v = mv1 t1 × v1t t1 = mv1 2 t t1 2 50 (b) If W1 = work done by applied force W2 = work done against friction then applying work energy theorem W1 − W2 = PE +KE (at the top) F × s − W2 = mgh + 1 2 mv 2 100 × 12 − W2 = 50 × 10 × 2 + 1 2 × 50 × 2 2 1200 − W2 = 1100 W2 = 100 J 51 (b) Fraction of length of the chain hanging from the table = 1 n = 60 cm 200 cm = 3 10 ⇒ n = 10 3 Work done in pulling the chain on the table W = mgL 2n 2 = 4 × 10 × 2 2 × (10/3) 2 = 3.6 J 52 (b) P = 3t 2 − 2t + 1 = dE dt ∴ dE = (3t 2 − 2t + 1)dt E = ∫ (3t 2 − 2t + 1)dt t=4s t=2s L/n