Title 05.WORK,ENERGY AND POWER.pdf ✅

Objective Physics Volume-I 683 YCT 05. Work, Energy and Power (a) Work done by Force and Power 1. A force acts on a 30 g particle in such a way that the position of the particle as a function of time is given by x = αt 2 , where x is in metre, t is in seconds and α = 1 m/s2 . The work done during the first 4 s is (a) 0.96 J (b) 0.45 J (c) 0.49 J (d) 0.53 J TS-EAMCET-10.09.2020, Shift-1 TS-EAMCET.11.09.2020, Shift-2 TS-EAMCET.14.09.2020, Shift-2 TS-EAMCET-05.05.2018, Shift-2 Ans. (a) : Given, mass of particle (m) = 30g =3×10–2 kg Displacement of the particle ( ) 2 x = αt , α = 1 m/s2 ∴Velocity ( ) ( ) d d 2 v t = 2 t dt dt = = α α x Acceleration ( ) ( ) dv d a 2 t = 2 dt dt = = α α ∴Force on the particle, 2 F ma 3 10 2 − = = × × α 2 6 10 N− = × [ 1] ∴α = ∴Work done during first 4 s, 4 0 4 2 0 W F d = 6 10 d − = × ∫ ∫ x x 4 2 0 4 2 0 6 10 d 6 10 vdt − − = × = × ∫ ∫ x 4 2 0 4 2 –2 0 6 10 2 t dt t 6 10 2 . 2 − = × α   = × × α     ∫ 2 12 10 8 − = × × [ 1] ∵α = = 0.96 J 2. The power of a pump which can pump 200 kg of water to a height of 200 m in 10s is: (g = 10 m/s2 ) (a) 40 kW (b) 80 kW (c) 4000 kW (d) 960 kW CBSE PMT-2000 CPMT-1989 SCRA-1994 MH CET-2000 JIPMER-2002, 2001 AP EAMCET(Medical)-1998 Ans. (a) :Given,mass (m) = 200 kg, Height(h) = 200m Time (t) = 10 sec, (g) = 10 m/s2 We know that, power (P) = Work time So, the power of pump, P mgh t = P 200 10 200 10 × × = = 40000 W P = 40 kW 3. A body moves a distance of 10 m along a straight line under the action of a force of 5 N. If the work done is 25 J the angle which the force makes with the direction of motion of the body is (a) 0o (b) 30o (c) 60o (d) 90o Manipal UGET-2019 CG PET- 2006 NCERT-1980 JIPMER-1997 AIPMT-1999 BHU-2000 RPMT-2000 Odisha JEE-2002 Ans. (c): Given that, Displacement, d = 10m, Force, F = 5N, Work done, W = 25J We know that, W = F.d cosθ 25 = 5 × 10 × cosθ cosθ = 1 2 cosθ = cos60o θ = 60° Then, force make angle (θ) = 60° with direction of motion. 4. A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of 45o with the initial vertical direction is (a) Mg( 2 1) + (b) Mg 2 (c) Mg 2 (d) Mg( 2 1) − UP CPMT-2008 AIEEE-2008 JCECE-2008 UPSEE-2007
Objective Physics Volume-I 684 YCT Ans. (d) : Given that, mass = M kg, θ = 45o Let, length of string is l Work done is equal to gain in potential energy of mass W = F × l sin 45° = Fl 2 Change in potential energy = Mgh = Mg (l – l cos 45° ) = 1 Mg 1 2   −     l ∵ Work done = Change in potential energy F 1 Mg 1 2 l l 2   = −     F Mg 2 1 = − ( ) 5. 300 J of work is done in sliding a 2kg block up an inclined plane of height 10m. Taking g = 10 m/s2 , work done against friction is (a) 200 J (b) 100 J (c) Zero (d) 1000 J JIPMER-2008 MP PMT-2002 CBSE PMT-2006 AFMC-2006 Ans. (b) : Net work done = work done against gravitational force + work done against frictional force. W W W = +g f W W mgh = +f f 300 W 2 10 10 = + × × W 300 200 f = − Wf = 100 J 6. When a position dependent force F = 7 – 2x + 3x2 N acts on a small body of mass 2 kg and displaces it from x = 0 to x = 5 m, calculate the work done (in joules): (a) 70 (b) 270 (c) 35 (d) 135 AP EAMCET-23.09.2020, Shift-II Manipal UGET-2016 APMT -1992 Ans. (d) : Given, F = 7–2x+3x2 Work done is given by 2 1 W x 0 x dw F.dx = ∫ ∫ W ( ) 5 2 0 = + 7 – 2x 3x dx ∫ 5 2 3 0 2x 3x 7x 2 3   = − +     = 5 2 3 0   7x x x − +   = [7×5–52 +53 – 0] = 35–25+125 = 135 J 7. A body is initially at rest. It undergoes one- dimensional motion with constant acceleration. The power delivered to it at time t is proportional to : (a) 1/ 2 t (b) t (c) 3/ 2 t (d) 2 t Karnataka CET-2020 Assam CEE-2014 BCECE-2006 Ans. (b) : Suppose m = mass of the body a = acceleration produced in the body v = velocity of the body P = power delivered to the body in a time t v = u + at v = at [∵u = 0] Force, F = ma Power delivered to the body – P = F × v P = ma × at P = ma2 t So, P ∝ t 8. A uniform chain of mass M and length L is lying on a smooth horizontal table, with half of its length hanging down. The work done is pulling the entire chain up the table is (a) MgL 2 (b) MgL 4 (c) MgL 8 (d) MgL 16 AP EAMCET (23.09.2020) Shift-I BITSAT - 2009 AMU - 2004
Objective Physics Volume-I 685 YCT Ans. (c): Given, mass = M, Length= L mass of small part of chain = dm M dm dx L = work done to pulling (dm) mass dw = dm g.x M dw .dx g.x L   =     M dw gx dx L = By integrating L / 2 0 M dw g x dx L = ∫ 2 L / 2 0 M x W g L 2 = 2 M L / 4 W g L 2 = MgL W 8 = IInd Method– Centre of mass at point M As we know that, Work done (W) = Change in energy W = ∆E W = Mg ∆l ∆l= 1 L L 2 2 4 × = Now, as per given condition, W = M L × g × 2 4 W = MgL 8 9. A ball is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to (a) t 1/2 (b) t 3/4 (c) t 3/2 (d) t 2 AP EAMCET-24.09.2020, Shift-I BITSET - 2020 UPSEE - 2010 Ans. (c) : For moving of a body along the straight line, machine need to deliver a constant power– P = Force (F) × Velocity (v) ....(i) ∵ F = ma and ∴ v 1 s a t t t   = =     Acceleration, 2 s a ,velocity, v at t = = (∵ Initial velocity, u = 0) From, eqn . (i), P = ma × v P = ma at × P = ma2 t 2 3 ms P t = 3 2 Pt ms = For constant power and mass distance will be– 2 3 s t ∝ 3/ 2 s t ∝ Where, s = distance moved by body t = time 10. If a force F is applied on a body and it moves with a velocity v, the power will be (a) F.v (b) F v (c) 2 F v (d) F.v2 AP EAMCET (18.09.2020) Shift-I UPSEE - 2004 AIIMS -1998 Ans. (a) : Power = Force × Velocity P = F.v Power can be define as a scalar product of force and velocity. 11. The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is S(in m) (a) 225 J (b) 200 J (c) 400 J (d) 175 J JIPMER-2012 Manipal UGET-2009 Karnataka CET-2009
Objective Physics Volume-I 686 YCT Ans. (b): Work done is equal to the area under the graph– W = area of ABC + Area of BFGC + area of DEF = ( ) 1 1 × 5 ×10 + 15 ×10 + × 5 ×10 2 2             = 25 + 150 + 25 = 200 J 12. A body of mass 3 kg in under a force which caused displacement in it, given by 2 t s = 3 in meter with time t in seconds. What is the work done by the force between time t = 0 and t = 2 is (a) 8 J (b) 5.2 J (c) 3.9 J (d) 2.6 J SCRA-2009 WB JEE-2008 AIPMT-2006 Ans. (d) : Given, Distance, 2 t s 3 = ds 2t dt 3 = 2t ds dt 3 = We know that- ds 2t v dt 3 = = Also, dv a dt = ∴ d 2t 2 a dt 3 3   = =     W F.ds F.ds = = ∫ ∫ W ma.ds = ∫ 2 0 2 2t W 3 dt 3 3 = × × ∫ 2 2 2 0 0 4 4 t W tdt 3 3 2   = =     ∫ 4 16 W [4] 6 6 = = W = 2.6 J 13. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s. The work done by the force of gravity during the time the particle goes up is (a) –0.5 J (b) – 1.25 J (c) 1.25 J (d) 0.5 J JCECE-2008 AIEEE-2008 UPSEE – 2007 Ans. (b) : Given, mass (m) = 100 g = 0.1 kg, u = 5 m/s v = 0 m/s From third equation of motion– v 2 = u 2 + 2gh 0 = u2 – 2gh u 2 = 2gh h = 2 u 2g = ( )2 5 25 2 9.8 2 9.8 = × × Work done by gravity (W) = mgh cos θ W = 0.1 × 9.8 × ( ) 25 cos180 180 2 9.8 ° θ = ° × ∵ ∴ W = – 0.1 × 25 –1.25J 2 = 14. An engine pumps up 100 kg of water through a height of 10m in 5s. Given that the efficiency of engine is 60%. If g = 10 ms–2, the power of the engine is– (a) 3.3 kW (b) 0.33 kW (c) 0.033 kW (d) 33 kW JIPMER -2014 BCECE-2008 UP CPMT-2006 Ans. (a) : Given, m = 100 kg, h = 10 m, t = 5 sec, g = 10 ms–2 Efficiency (η) = 60% = 0.6 Work done = mgh W = 100 × 10 × 10 = 10000 J Effective power = Work Time = 10000 5 = 2000 watt Effective power = Efficiency × Total power Total power = Effective power Efficiency Total power = 2000 0.6 = 3333 watt = 3.33 kW 15. A force F (5i 4j)N ˆ ˆ = + acts on a body and produces a displacement S (6i 5j 3k)m. ˆ ˆ ˆ = − + The work done by the force is (a) 10 J (b) 20 J (c) 30 J (d) 40 J AP EAMCET (17.09.2020) Shift-II Manipal UGET-2013