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WORK-ENERGY-POWER

WORK-ENERGY-POWER. 1. (i) WORK :- Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity. Unit: Nm ( Joule ). F. F. α. α. A. B. s.

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WORK-ENERGY-POWER

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  1. WORK-ENERGY-POWER 1 (i)WORK:- Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity. Unit: Nm ( Joule ) F F α α A B s Work done=F × cosα× s α= angle of inclination of force with the direction of motion s=displacement of force from A to B

  2. 2 (ii)POWER:- It is defined as the time rate of doing work. Power = work done /Time= (force × distance) /Time   = force × velocity Unit: (Nm)/s [watt] (kN m)/s [kilo watt] 1 metric H.P=735.75 watts.

  3. 3 • Energy:- • It is defined as the capacity to do work. It is a scalar quantity. • Unit :- N m (Joule)

  4. Work-Energy relation for translation 4 • From Newton’s second law of motion • ∑ F =(W/g) × a -----------(1) • Also a = dv/dt =( dv/ds) ×( ds/dt) = v × dv/ds • sub in (1) • ∑ F=( W/g ) × v × dv/ds • F × ds = (W/g ) × v × dv ------------------(2) • Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v

  5. 5 Integrating both sides, we get s v  F  ds = (W/g) v dv 0 u V F × s = ( W/g ) v2/2] u ∑ F × s = (W/2g) [v2-u2] Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.

  6. 6 Work done by a spring s F=k(s) The force required to cause unit deformation of the spring is called the spring modulus denoted by the symbol k. The force required to deform a spring is given by F= ks. Work done by the force on a spring is the product of the average force and the deformation s. W= -ks2/2. [ The negative sign indicates that whenever spring is deformed the force of spring is in the opposite direction of deformation.]

  7. IMPULSE – MOMENTUM 7 Momentum:- Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity. Unit:- N s. Impulse of a Force:- It is defined as the product of force and the time over which it acts. It is a vector quantity. Unit:- N s.

  8. 8 Impulse-momentum relationship Force = Rate of change of momentum F =( mv – mu) /t = Force causing impulse  F×t = mv – mu Impulse = final momentum – Initial momentum The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.

  9. 9 Law of Conservation of momentum The law of conservation of momentum may be stated as , “momentum is conserved in a system in which resultant force is zero”. In other words, in a system if the resultant is zero, Initial momentum is equal to Final momentum m1u1+m2u2 = m1v1+m2v2

  10. 10 PROBLEMS 1.     A block weighing 2500N rests on a level horizontal floor for which  = 0.2. This block is pulled by a force of 1000N at an angle of 30o to horizontal.Find the velocity of block after it moves 30m starting from rest. If the force of 1000N is then removed, how much further will the block move. 2500N 1000N 30º

  11. 11 FBD OF BLOCK Solution: Case (i) RN – 2500 + 1000sin30o = 0 RN = 2000N from W.E. principle  F×s = (W/2g) (v2 – u2) (1000×cos30o – 0.2 × 2000) 30 =[2500 /(2×9.81)] [(v2 – 0)]. v=10.47m/s 2500N 1000N 30º F=0.2×RN RN=2000N

  12. 12 Case (ii) Force of 1000N removed 2500N 10.47m/s F=0.2×RN RN = 2500N u = 10.47m/s, v = 0, s = ?  F×s= [(W/2g) (v2 – u2)] ( – 0.2 ×2500) s = [2500/(2×9.81)] [0 – (10.47)2] s = 27.94m

  13. 13 2.     Determine using work energy principle the velocity of block shown in figure after it has moved 50m starting from rest. Assume  = 0.2. 4000N 30º 20º 5000N

  14. 14 F.B.D Solution: RN – 5000cos20º + 4000sin30º = 0 RN = 2698.46N from W.E. principle  F×s=(W/2g) (v2 – u2) (4000cos30º – 539.69 – 5000sin20º)50 = 5000 ( v2 – 0 ) 2×9.81 v = 15.44 m/sec. 4000N 30º F=0.2×RN =539.69N RN 20º 5000N

  15. 3.Determine the constant force P that will give the system of bodies shown in figure a velocity of 3m/s. after moving 4.5m from rest. The pulleys are frictionless and  = 0.2.WA=200N, WB=1000N, WC=500N. 15 P A B 4 3 C

  16. 16 200N From W.E. principle  F×s = [W/2g] (v2– u2) [P – 40 – 120 – 1000 sin 53.13o – 100] 4.5 = 1700 [(3.0)2 – 0 ] 2 ×9.81 P = 1233.29N P 0.2×200=40N RN=200N 0.2×600=120N 500N =600N RN=1000×cos53.13º 53.13º 1000N 0.2×500=100N RN= 500N

  17. 17 4. In what distance will block A attain a velocity of 3m/s starting from rest? Take  = 0.2. WA = 1500N, WB = 2000N 3 A 4 B 4 3

  18. 18 SOLUTION : For the arrangement shown in figure the displacement and velocity of (B) are one half of corresponding value of (A). Also the internal forces (tensions in cords) do no work F B D from W.E. principle  F×S = W/2g (v2-u2) (1500 × 4/5-180) sA + ( - 2000×3/5- 320) sA/2= [1500/(2×9.81)](3)2 +[2000/(2×9.81)](1.5)2 sA = 3.53m. A F1=0.2×900 =180N B RN1= 1500×3/5=900N RN2=2000×4/5 =1600N 1500N F2=0.2×1600=320N 2000N

  19. 19 5.An engine weighing 500 kN drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same ? What is the tension developed in the link connecting the engine and carriages? 500N P 1500N 100 1

  20. 20 F.B.D P F1=2.5kN F2=7.5kN 500kN 1500kN Track resistance = 5N/kN Engine ----- 5 ×500 = 2500N = 2.5 kN Carriages ----- 5 × 1500 = 7500N = 7.5 kN from W.E. principle  F×s = (W/2g) (v2 – u2) [P – 2.5 – 500 × (1/100)– 7.5 – 1500 ×(1/100)] 1000 = [2000/(2×9.81)][(10)2-0] P = 40.19 kN Power developed = 40.19 × 10 = 401.9kW

  21. 21 F.B.D. of carriage T W.E. Principle  F×s= (W/2g) (v2 – u2) [T – 7.5 – 1500×1/100] 1000 = 1500/(2×9.81) [(10)2 -0] T = 30.15 KN F=7.5kN 1500N

  22. 22 6.A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph (i) Up an incline 1 in 300 (ii) Down an incline 1 in 250. Take efficiency 75%. Case (i) Up the incline 20kmph = 5.56m/s for uniform speed Σ F = 0 P – 1.5 – 150 ×1/300 = 0 P = 2kN Output Power = 2 × 5.56 = 11.12 kW. Input power = 11.12 = 14.81 kW 0.75 P F=1/100×150=1.5kN 1in300 150kN

  23. 23 Case (ii) Down the incline Σ F = 0 P + 150 ×1/250 – 1.5 = 0 P = 0.9kN Output Power = 5.56×0.9 = 5kW. Input power = 5/0.75 = 6.67 kW F=1.5kN P 1in250 150kN

  24. 24 7.Block ‘A’ of weight 50 N is released from rest from the position shown in figure. Determine the maximum compression of the spring .The Modulus of spring is 800 N/m and  between block and plane is 0.2. Also find the maximum velocity of the block. 50N 0.8m 30º

  25. 25 800+s F=0.2×43.3 k=800N/m =0.8N/mm =8.66N 30º RN=50×cos30º 50N =43.3N Solution : Let ‘s’ mm be the maximum compression. The body was at rest and again at rest when it moves a distance of (800 + s) mm Apply W.E. principle (50 sin 30º – 8.66) (800 + s) – ½×k×s2 = 0 (50 sin 30º – 8.66) (800 + s) = ½ × 0.8 × s2 16.34 (800 + s) = 0.4× s2 13072 + 16.34 s = 0.4×s2 400s2 – 16.34 s– 13072 = 0 s = 202.35mm

  26. 26 800+x F=0.2×43.3 =8.66N 30º RN=43.3N 50N The velocity will be maximum when dv/dt = 0 and therefore net force acting on the body is zero; when spring force developed balances the force exerted by the body. Let x be the deformation when the net force on the body in the direction of motion is zero. kx = W sinθ – F 0.8 x = 50 sin 30º – 8.6, x = 20.4mm Apply W.E principle (50 Sin 30º – 8.66) (800 + x) – ½ kx2 = 50 [(vmax)2 – 0] 2 × 9810 (50 Sin 30º – 8.66) (800 + 20.4) – ½× 0.8× (20.4)2 = 50 (vmax)2 2×9810 vmax = 2279.23 mm/s=2.28m/s

  27. 27 8. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest. 3kg 5kg 30º 30º

  28. 28 A B 30º 30º 5×9.81=49.05N 3×9.81=29.43N Case (i) F.B.D 0f entire system From impulse momentum principle Σ F×t = mv – mu (49.05 Sin30º – 29.43 Sin30º) 2 = 8(v –0) v = 2.45 m/s.

  29. 29 2.45m/s 30º 29.43N Case (ii) F.B.D of Block B : From W.E. principle Σ F×s= (W/2g) (v2 – u2) ( – 29.43 Sin30ºs)= 29.43 [0-(2.45)2] 2 × 9.81 s= 0.61m

  30. 30 9.A locomotive weighing 900 kN pulls a train of 10 coaches each weighing 300 kN at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest. 900kN 6x300=1800kN 4x300=1200kN P

  31. 31 900kN 4×300=1200kN 6×300=1800kN P F3=8.4kN F2=12.6kN F1=6.3kN Solution : 72 kmph = 20m/sec F1 = 7×900 = 6300 N = 6.3 kN F2 = 7 ×1800 = 12.6 × 103 N = 12.6 kN F3 = 7×1200 = 8400N = 8.4kN Σ F = 0 P – 8.4 – 12.6 – 6.3 = 0 P = 27.3 kN Impulse momentum for engine and 6 coaches Σ F×t = mv – mu (27.3 – 12.6 – 6.3) 120 = 2700 ( v – 20 ) 9.81 v = 23.66 m/s

  32. 32 F.B.D OF DETACHED COACHES 4×300=1200kN 20m/s F3=8.4kN W.E. principle for snapped coaches Σ F×s=( W/2g) (v2 – u2) – 8.4×s = 1200 [ 0 – (20)2] 2 × 9.81 s = 2912.48m s = 2.9 km.

  33. 33 10.Determine the velocity of block B after moving for 5 sec. starting from (i) rest (ii) downward velocity of 2 m/s. BLOCK A=450N BLOCK B=650N A B

  34. 34 FBD of B FBD of A Case (i) Starting from rest: From impulse momentum principle Σ F×t = mv – mu Block B(2T – 650) 5 = 650 ( vB – 0) 9.81 T – 325 = 6.63 vB ---- (1) Block A (450 – T) 5 = 450 ( 2vB – 0) 9.81 450 – T = 18.35 vB--- (2) Solving 1 and 2 vB = 5 m/s. 2T T uB=0 vB uA=0 vA=2vB 650N 450N

  35. FBD of B 35 2T Case (ii) Downward velocity of 2m/s Σ F×t = mv – mu Block B (2T – 650) 5 = 650 (vB +2) 9.81 T – 338.25 = 6.63 vB ---- (1) Block A(450 – T) 5 = 450 (2vB +4) 9.81 413.3 – T = 18.35 vB ----- (2) Solving 1 and 2 vB = 3 m/s. uB = -2m/s vB 650N FBD of A T uA= 2(-2) = - 4m/s vA= 2vB 450N

  36. 36 11.  A 50 Kg man and 25 Kg boy jump to a pier from a 300 Kg row boat with a velocity of 5m/s relative to the boat. If the boat is initially at rest in water, in which of the following conditions will the boat acquire largest velocity. (a)   Man and boy jump simultaneously (b)  Man jumps first, followed by boy (c)  Boy jumps first, followed by man. Solution : If vm ,vb= Velocity of man, Velocity of boy v= Velocity of boat 5 = vm – (– v) vm = 5 – v or vb = 5 – v

  37. 37 Case (i) Both jump simultaneously:- u=0 v 5 - v = + m=375kg m1=300kg m2=75kg From Law of conservation of momentum 375×0 = -300v + 75( 5 – v) 0 = -300v + 375 – 75v 375v = 375, v = 1m/sec

  38. Case (ii) Man jumps first followed by boy:- 38 v1 u = 0 5 – v1 = + m=375kg m1=325kg m2=50kg 375 ×0 = -325×v1 + 50 ( 5 – v1) v1 = 0.67 m/sec 0.67m/s v 5 – v = + m1=300kg m=325kg m2=25kg -325×0 .67 = -300× v + 25 ( 5 – v) v= 1.05 m/sec

  39. Case (iii) Boy jumps first followed by man:- 39 u = 0 v1 5 – v1 = + m=375kg m1=350kg m2=25kg 375 ×0 = -350v1 + 25( 5 – v1) v1 = 0.33 m/sec 0.33m/s v 5 – v = + m1=300kg m=350kg m2=50kg -350 × 0 .33 = -300×v + 50( 5 – v) v= 1.047 m/sec Largest Velocity v = 1.05 m/s

  40. 40 12. A man weighing 800N moves horizontally with a velocity of 3m/s and jumps on to a boat which is at rest and at the same level. If the boat weighs 3200N, what is the velocity of both? What is the distance traveled before coming to rest if they are subjected to an opposing force of 50N? Part (ii) :Distance traveled From W.E. Principle  F×s = (w /2g) (v2 –u2) -50 s = [407.75/(2×9.81)] [ 0 – (0.6)2 ] s = 1.47 m Part (i): Velocity of man & boat m1 = 800/9.81 = 81.55kg. m2 = 3200/9.81 = 326.2 kg m1u1+ m2u2 = (m1+m2 ) v (81.55 × 3 ) + 0 = (407.75) v  v = 0.6 m/s

  41. 41 13. A pile hammer weighing 20 KN drops from a height of 750mm on a pile of 10 KN. The pile penetrates 100mm per blow. Assuming the motion of pile is resisted by a constant force, find the resistance to penetration offered by the soil. Velocity of hammer just before it hits the pile u1 = ( 2gh) = ( 2 × 9.81 × 0.75) u1 = 3.84 m/s From law of conservation of momentum m1u1 + m2u2 = (m1 + m2) v (20/9.81) (3.84) + (10×0) = (30/9.81) v v= 2.56 m/s From W.E. Principle  F×S = (w/2g) [(v2 – u2)] ( -R + 30) 0.1= [30 / (2 × 9.81)] [0 – (2.56)2]   R = 70.21 KN 20KN 0.75m 10KN 0.1m

  42. 42 14. A bullet of weight 0.3N is fired horizontally into a body of weight 100N which is suspended by a string 0.8m long. Due to this impact the body swings through an angle of 30o. Find the velocity of the bullet. Cos 30 = (0.8 – h) / 0.8 h = 0.107m From law of conservation of momentum m1u1+m2u2 = (m1+ m2 ) v (0.3/9.81) u1 + 0 = [ (0.3 + 100)/ 9.81 ] v--(1) from W.E .Principle  F×S= (w/2g) (v2 – u2) 100.3 (-0.107) = [100.3/(2 × 9.81)] (0 – u2) u = 1.45m/s u = v = 1.45 m/s in equation (1) sub. In (1) u1 = 493.97 m/s 30° 0.8 - h 0.8m + h +

  43. 43 15.A body of weight 500N slides up an incline of 30o under the action of an applied force of 800N along the incline. If the body moves from rest, determine the following (i) velocity after 4 secs. (ii) K.E. (iii) impulse (iv) momentum If the body moves now along level surface and with the external force removed, how long will it take for the body to come to rest and how far will it have traveled?( Take  = 0.2)

  44. 44 (i)     F x t = mv –mu (800 – 86.6 – 500 Sin30) 4 = 500/9.81 (V- 0) V = 36.36m/s FBD 800N ii)  K.E. = ½ mv2 = ½ × (500 / 9.81) × (36.36)2 = 33691.38 N m iii) impulse =  F× t = (800 – 86.86 – 500Sin30) 4 = 1852.56 N s iv) Momentum = m × v =( 500/ 9.81) × 36.36 = 1853.21 N s  1852.56 N s F = 0.2 × 433.01 = 86.6N 30° RN = 500Cos30° = 433.01N 500N

  45. 45 v=0 500N s u=36.36m/s F=0.2×500=100N RN=500N  F× t = mv –mu -100 t = 500/9.81 (0 – 36.36) t = 18.53 sec.  F×s = (W/2g) (v2 – u2) -100 s = [500/(2 × 9.81)] [ 0 – (36.36)2 ] s = 336.91m

  46. 46 ASSIGNMENT 1.A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel before coming to rest . Take µk=0.3 [Ans :s=6m ] 5m A 3 C 4 B s

  47. 47 2.The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m] A B 3 4 3m

  48. 3.Two bodies a and b weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a after the block b hits block c .[ Answer s=1.34m] 48 A 5 B 3m 12 4.A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.[ ANSWER: power developed = 6.131KN]

  49. 49 5.A spring is used to stop 60kg pack age which is sliding on a horizontal surface . the spring has a constant k which is 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the co-eff of kinetic friction between package and surface. (Answer µk=0.2) 2.5 m/s 60kg 600m

  50. 50 6. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N] P 1000N 400N

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