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Topics in textbook Actual Work Definition of Work Work on Particle Potential Energy Definition of Energy Work-Energy Equation Potential Energy Kinetics Energy Definition of Energs 1 Three approaches for solving dynamics 1) Direct Method 1 m From 2nd Law (kinetics Eq) Fi dr m(a dr ) i i Fi Fi (a, v , s , t ) Kinematics Eq: a (v , s , t ) v , s F2 path dr F a A F3 F1 work (and potential energy) of Force i along the path 2) Work and Energy From 2nd Law F ma Newton’s 2nd Law i Fi dr m(a dr ) s m(a dr ) m(v dv ) t t 1 1 mv2 2 mv2 2 2 2 (change of kinetics energy) F s i i dr s m(a dr ) 3) Impulse and Momentum From 2nd Law F s i i dt s m(a dt ) i m(a ) dt m(dv ) d (mv ) mv mv F dt m ( a ) dt i linear impulse of Force i 2 (linear momentum) 1 2 F ma Newton’s Work and Energy From 2nd Law 2nd Law Usually convenient when F = F(s), and you want to find velocity at final state (without finding acc. first). change of kinetic energy i Fi dr m(a dr ) s def Ui F dr t m(v t dvt ) s Work of Force i along the path Fi dr Fi cosi ds Fi (ds)cosi dU1 0 dU2 0 dU3 0 1 1 mvB 2 mv A 2 2 2 TB TA TB A during small displacement dUi Fi kinetic energy at B m(a dr ) m(a ds) i Work of a force i T 1 2 mv 2 kinetic energy F2 path 1 kinetic energy at A dr A B F3 F1 Principle of work and Energy U AB T TB TA 3 Work done on Particle dU1 0 Work by a force Fi dUi Fi dr Fi (dr )cosi i F3 F1 path (Fi ) (dr ) cosi dr ( F )cosi dU2 0 dU3 0 (Fi ) cosi (dr ) P F2 (inactive force) Work done over particle A dU (F i sum of works done by all forces over the particle A i dr ) ( Fi ) dr Since F ma , the total work done on object is 5 …… Note on work F2 path dr dUi Fi dr F3 F1 dU i is positive when Fi t F (cosi ) and s has the same direction. Unit of work is N-m or Joule (J). Active force is the force that does the work Reactive force = constrain force that does not do the work 6 B The 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m, determine the total work done by all forces acting on the block when a horizontal force P = 400 N pushes the block up the plane s = 2 m. The block is not tipping. A (spring stretched length is 0.5m) Pos B Horizontal Force P: constant Pos A UP (P cos30 )(2) 692.8J o Weight W: constant. UW mg (2sin30 ) 98.1J “Active Force” Spring Force Fs.: varying Force 2 2 U Fs ( Fs )(dx) k ( x 0.5)dx 1 k ( x 0.5) 2 90J A 2 0 0 B Normal Force NB : constant U NB 0 “Inactive Force” U Overall 692.8 98.1 90 504.7 J Work done on Particle t Work done over particle A F4 path dr F3 a F1 dU (F dr ) ( F ) dr sum of works done by all forces i i P (ma ) dr over the particle A F2 B B B A A A U AB dU m a dr m at ds Pos B TB Pos A i B B TA Work done on particle P during path A->B, is to increase kinetic energy of particle 1 m vdv mv 2 1 mvB2 1 mvA2 2 2 2 A A def 1 T mv 2 2 TB TA Kinetics Energy 1 2 T mv 2 path F2 A T is the work done on a particle to accelerate it from rest to the velocity v F3 F1 Principle of work and Energy Unit of T is N-m or Joule (J) U AB TB TA Advantage ( TA U AB TB ) Scalar equation. (1 unknown) Integral Equation (not instantaneous eq like 2nd Law) No need to find acceleration first Get change in velocity directly from active forces. it can be applied to system of particles with frictionless and non-deformable links 12 15 18 How to calculate Work F2 dr In general F3 U i dU i Fi dr s s In xyz-coord Ui path 1 Fi F Fi x iˆ Fi y ˆj Fi z kˆ dr (dx)iˆ (dy) ˆj (dz)kˆ Fi x dx Fi y dy Fi z dz make sure that the direction of Fi x In nt-coord Ui scalar (be careful Of +/-) Fi y Fi z is positive according to x y z direction. Fi t ds In r-coord Ui dUi Fi dr Fi cos i ds displacement x force component in the direction of displacement Fi (ds)cos i displacement in the direction of force x force component make sure that the direction of Fi t is positive according to +s direction. F Fr eˆr F eˆ Fr dr F rd dr (dr )eˆr (rd )eˆ 19 M3/107) Calculate the work done on 10-kg object with the constant Force ( F= 8N ) during the curve path AB. F 8iˆ (const) x-y dr dx iˆ dy ˆj U F , A B F dr 0.75 8 dx 8(0.75) J 0 mg mg ˆj U mg , A B mg dr y 0 mg dy 0.375 mg (0.375) x mg F=8N (const) N Umg , AB (mg )(0.375) 36.7875 J U N , A B 0 Does not do the work U AB U F , AB U mg , AB U N , AB 42.7875 J If F is not constant, how to calculate it? Ans If F is not constant U F , A B F dr F ( x y)iˆ ( x2 y 2 ) ˆj B ( x y) dx ( x 2 y 2 ) dy A y B ( x (0.375 0.667 x 2 ) dx y 0.375 0.667 x A 2 0.375 y 2 y dy 0.667 A B x More general case F Fx ( x, y)iˆ Fy ( x, y) ˆj U F , A B F dr B Fx dx Fy dy A B Fx ( x, f ( x)) dx Fy ( x, f ( x)) dy A y f ( x) Fy ( f 1 ( y), f ( y))dy or Fy ( x, f ( x)) dy dx 21 dx M3/107) Calculate the work done by F during the curve path AB. n-t s F (s) Ft (s)eˆt Fn ( s)eˆn Fn ( s ) Fn doest not effect works! Ft 2s cos Fn 2s sin engine thrust U F , A B F dr Ft ( s )ds F (t ) 2s 2 s cos ds M3/107) Calculate the work done by F during the curve path AB. r- FC eˆr F FCC eˆr U F , A B F dr rA rB F (const) r- reference point d 0 (dr )eˆr rd eˆ central force Feˆr 0eˆ (dr )eˆr (rd )eˆ B FC dr FC A r reˆr FC rB rA dr dr eˆr r d eˆ dr eˆr r eˆ B dr A FC (rA rB ) 23 M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B in the vertical plane under the action of the constant 5-N tension in the cord. If the slider is released from rest at A, calculate its velocity v as it reaches B. mg F Work-Energy Eq. 0 general position U AB 12 mv TB T A N 2 B Does not do the work eˆr F r- coordinate reference point Umg mg (0.25) 0.4905 J B U F Fdr A F (rA rB ) F rB dr F (r B rA ) rA 5( 0.62 0.252 0.15) 2.5 vB 2U A B m 2 2.0095 4.482745 0.2 What does it mean if U A B 0? U AB U F Umg U N 0 2.5 0.4905 2.0095 J 25 Work on frictionless connected particles Only the external forces are needed to calculate the total work on a system of particles. (If frictions exist, the sum of action and reaction of the friction may not be zero.) internal force R and –R will have the same displacement. So, the sum of these works are zero. 33 The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant) initial state Final state mA g mA g mB g mB g U A 12 TA 0.5 Fd mA gd Tdy TA 0 UB 12 TB N causes no work! 0.5 mB gd Tdy TB 0 System selection is not so good (you have to calculate Tension T for its work) U sys mA g mB g 1 2 TA Fd mA gd mB gd TA 1 1 mAv 2 mB v 2 2 2 34 M3/131) The ball is released from position A with a velocity of 3m/s and swings in a vertical plane. At the bottom position, the cord strikes the fixed bar at B, and the ball continues to swing in the dashed arc. Calcuate the velocity v of the ball as it passes position C. Work-Energy Eq. U AC 12TmvC 12TmvA 2 C T U AC U mg AC 2 A UT AC does no work system U mg mg (hA hB ) mg (1.2cos 60o 0.8) mg (0.2) mg (0.2) mg vC 2 1 2 2 U mv 12.924 3.595 m/s A mg m 2 35 F2 Power path dr Power is defined as time rate of work d F r dU P F v dt dt A F3 F1 (scalar quantity) For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) Unit of power: Watt (W) = J/s = N-m/s 37 Mechanical Efficiency Mechanical Efficiency useful power output power input energy output energy input 1 If energy applied to the machine occurs during the same time interval at which it is removed. Poutput F v Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions. 38 A car has a mass of 2 Mg and an engine efficiency of = 0.65. The car uniformly accelerates at 5 m/s2, starting from rest. During that constant acceleration, the wind outside creates a drag resistance on the car of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output input when t=4 s. a Poutput Poutput F v Pinput Poutput t 4 F t 4 v t 4 x 10480 20 209.6 kW Pinput [ : Fx max ] Constant acceleration: t 4 Poutput 209.6 322.46 kW 0.65 FC 1.2v2 ma FC ma 1.2v2 10480 N a 5 (constant) v at v t 4 (5)(4) 20 m/s 39 A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. P v=? sP F = 30 N (const) output Pinput Pinput LM 2sP C 50N s = 10 m (start from rest) dsP d vM LM 2 2vP 2vB dt dt Energy F ma Approach B TBma TA 2UFAB 50 1 (2 F2)(Fs) 50mgs mv 2 a 9.812 v 50 N 375.6 494.2 W 0.76 (30)(2 6.26) 375.6 W Poutput F v 2F= 2(30) Poutput 5029.81 2 Fs mgs v u 2as 2 2 m 6.26 v 2(9.81)10 Work and Energy kinetic energy at B 1 1 2 mv mvA2 B 2 2 U AB TB TA F1 kinetic energy at A path dr B A F2 F3 summation of all forces UF1 , AB Work from all other forces (not spring & gravitation) UF2 , AB Work from Gravity Force Gravitational Potential Energy UF3 , AB Work from spring Elastic Potential Energy We found that …. It is much easier to solve dynamic problem, if we think the work done by spring and gravity force in the form of Potential Energy 42 Work of Gravity Force dUW W (dh) >0 UW ,12 mg (h1 h2 ) Vg ,1 Only depends on position at final state (2) Only depends on position at initial state (2) any path 1 energy level (higher) W=mg Vg ,2 2 h1 h mg (h2 h1 ) (mgh2 mgh1 ) Work done by W , only depends on the initial state position and final state position only, i.e. , it does not depends on actual path energy level (lower) h2 Think in Term of “Potential Energy” (for convenience) Fixed reference line UW , anyPath 12 (Vg ,2 Vg ,1 ) Work done by Gravity Force: from position 1 to position 2 U V Work = “Energy in Transfer” def Vg mgh point function Potential Energy - Energy from gravity field Vg ,2 Vg ,1 : Change in Internal Energy: + (increasing) V Energy “Emission”: from position 1 to position 2 h1 h2 Vg ,1 Vg ,2 when change in g is significant Define Vg as negative of work done from the position to r GM earth F m 2 r g GM earth Rearth 2 9.81 mgR 2 mgR 2 Vg (r ) 2 dr r r r the potential energy at r is mg Rearth Vg r 2 Vg from r1 to r2 Vg GM earth 0 Rearth mgR 2 mgR 2 Vg r2 r1 44 Work of Spring Force E1 Lo L1 Lo L2 L1 1 Ve,1 Ve,2 L dU Fs Fs dL Only depends on position at initial state (1) 2 U Fs ,12 k ( Lo L)dL 1 L2 L2 1 1 1 2 k ( L Lo )2 k ( L L ) k ( L1 Lo )2 2 o 2 L1 2 2 Lo natural length (unstretched length) E2 Fs Only depends on position at final state (2) Work done by Spring , depends only on the initial state and final state only, i.e. , it does not depends on actual path 2 any path def Think in Term of “Energy” (for convenience) U Fs , anyPath 12 1 Ve kx 2 2 x : distance , stretched or compressed from natural length point function (Ve,2 Ve,1 ) (Ve,1 Ve,2 ) Work done by Spring Force: from position 1 to position 2 U V Energy Emission: from position 1 to position 2 Work-Energy Equation FBD ** (Use Energy Concept) FBD Not Recommended Method in this course N N Work-Energy Equation Work-Energy Equation (1st Form) U AB T U AB U A B T * Virtual work by nonconservative forces. U A B V VB VA Energy Concept (2nd Form) U *AB V T Work-Energy Equation (1st Form) U AB T FBD Work-Energy Equation (2nd Form) U AB V T * N = E M3/173) The 0.6-kg slider is released from rest at A and slides down under the influence of its own weight and of the spring of k = 120 N/m. Determine the speed of the slider and the normal force at point B. The U U 0 unstrecthed length of the spring is 200 mm. * AB N , AB U A* B V T F N y 2x VA TA VB TB mg 5.866 2 1 1 1 1 mghA kxA 2 mvA2 mghB kxB 2 mvB2 2 2 2 2 gravitational potential datum hA 0.5 x A ( 0.52 0.252 0.2) xB (0.25 0.2) vB 5.9234 m/s At position B FB k (0.25 0.2) man mg mat NB NB mg Fs m vB2 an N B 84.09 N 3 2 2 df 1 dx 2 d y dx 2 1 (4 x) 2 4 3 2 x 0 1 m 4 49 Advantage Integral Equation (not instantaneous equation like 2nd Law) Scalar equation. (easy to handle with1 unknown) Get change in velocity directly. (No need to find acceleration first) Handle with only active forces. it can be applied to system of particles with frictionless and non-deformable links We will see this later, when applying at system of particles 57 Work on frictionless connected particles A B C O 58 initial state The system starts from rest at Configuration 1. Find the velocity of A at configuration where d = 0.5 m . F is 20 N (constant) Final state T T mA g T mB g T F mA g mB g F Object A 0.5 U A* ,12 VA TA Fd Tdy VA TA 0 Object B Ny U B* ,12 VA TA Nx * Usys ,12 Vsys Tsys F 0.5 Tdy VA TA 0 Fd VA TA T is internal force (excluding from Work Calculation) We have no interest in T, thus object separation (separating object A and B) is not good in this problem. 60 64 M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. SA SB System: block A + block B + cord+ 2 Pulleys Position A: at rest (assume) up Position B: block B moves down as 1 meter U * sys , AB V T 2 2 sB 3 3 2 hA sin 20o 3 3vA 2vB 0 3aA 2aB 0 datum 1 1 mA g hA mB g hB mAvA2 0 mB vB2 0 0 2 2 s A L 3S A 2SB C1 sB 1 hB 1 2 v A vB 3 2 1 2 1 10.9924 mA vB mB vB2 0 2 3 2 unsolvable M3/158) If the system is released from rest, determine the speeds of both masses after B have move 1 m. Neglect friction and the masses of pulleys. SA SB System: block A + block B + cord+ 2 Pulleys Position A: at rest (assume) up Position B: block B moves down as 1 meter U * sys , AB L 3S A 2SB C1 3vA 2vB 0 3aA 2aB 0 V T datum 1 1 mA g hA mB g hB mAvA2 0 mB vB2 0 0 2 2 2 2 sB 3 3 2 hA sin 20o 3 s A sB 1 hB 1 2 1 2 1 10.9924 mA vB mBvB2 0 2 3 2 2 v A vB 3 vB2 0.85285 vB 0.85285 vB 0.85285 vA 0.61566 H14/16) Block A rest on a surface which has friction. Determine the distance d cylinder B must move down so that A has a speed of vA 2 m/s starting from rest. sA 20 N k 0.3 sB System: block A + block B + cord+ 2 Pulleys 50 N Position A: at rest Position B: block B moves down as d meter * USys , AB E * USys , AB ( k N )(2d ) 1 1 2 E mA vA mB vB 2 (mB g )d 2 2 k N 2d 1 1 mA (2) 2 mB (1) 2 mB g (d ) 2 2 1 (4mA mB ) d 0.1744 m 2 (mB g 2k mA g ) L sA 2sB 0 sA 2sB sB : d sA 2d vA 2vB 0 vA : 2 vB 1 67 3/168) The system is released from rest with =180, where the uncompressed spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar. Determine the angle corresponding to the maximum spring compression. O2-1 * Usys , AB V T O2-2 O1 r L datum r VA TA VB TB /2 /2 1 1 m1 g (h1, A ) 2m2 g (h2, A ) kx A2 m1 g (h1, B ) 2m2 g (h2, B ) kxB2 2 2 /2 System: O1+O2+O3+4 rods Position A: at rest with =180 h1, A 180 2(0.2)sin 0.4 2 h2, A 180 (0.2 0.3)sin 0.5 2 Position B: maximum compression * (v* 0 for all object) h1 2r sin * h2, B (0.2 0.3)sin 2 * xB 2(0.2)sin 2(0.2) 2 * * 1 * 4(9.81)(0.4) 1 sin (2)(3)(9.81)(0.5) 1 sin 900 0.4 1 sin 2 2 2 2 2 h2 (r L) sin * h1, B 2(0.2) sin 2 2 * * 45.13 1 sin 0 or 1 sin 2 2 72 * 0 or 43.8o 2 68 Power F2 path dr Power is defined as time rate of work d F r dU P F v dt dt A F3 F1 (scalar quantity) For a machine, power tells how much work it can do in a period of time. (small machine can deliver lots of energy given enough time) Unit of power: Watt (W) = J/s = N-m/s 70 Mechanical Efficiency Mechanical Efficiency useful power output power input energy output energy input 1 If energy applied to the machine occurs during the same time interval at which it is removed. Poutput F v Since machine consists of moving parts which may have frictions, so extra energy or power is needed to overcome the frictions. 71 A 50-N load (B) is lifted up by the motor from rest until the distance is 10 m. The motor M has an efficiency of 0.76 and exerts a constant force of 30 N. Find the power supplied to the motor at that instant. Neglect the mass of the pulleys and cable. v=? sP F = 30 N (const) Poutput Pinput Pinput Poutput F v 375.6 494.2 W 0.76 ( 0.76) Poutput ? (30)(6.26) No! (30)(2 6.26) 375.6 W LM 2sP C 50N 2F= 2(30) Poutput Ny Nx s = 10 m (start from rest) vM Energy F ma Approach F = 30 N (const) ds d LM 2 P 2vP 2vB dt dt U2A*F (VBma VA ) (TB TA ) B50 1 2 2 F 50 a( F )(2 20) mgs 9.81 2 mv 50 9.81 v 50 N 50 N v u 2as 2 2 2 2 Fs mgs 6.26 v m 2(9.81)10 Summary Make sure you write FBD (no FBD, no score) * AB V T U AB T U Scalar Equation (Only 1 unknown) or Equation itself is not hard to solve, but calculating work may be more difficult than you thought. 74 Recommended Problem M3/155 M3/144 M3/166 M3/160 M3/168 H14/93 , H 14/96 75 77