Energy, Kinetic Energy, Work, Dot Product, and Power 8.01t Oct 13, 2004Energy Transformations • Falling water releases stored ‘gravitational potential energy’ turning into a ‘kinetic energy’of motion. • Human beings transform the stored chemical energy of food into ‘catabolic energy • Burning gasoline in car engines converts ‘chemical energy’ stored in the atomic bonds ofthe constituent atoms of gasoline into heatEnergy Transformations • Stretching or compressing a spring stores ‘elastic potential energy’ that can be released as kinetic energy • kinetic energy, gravitational energy, heat energy, elastic energy, electrical energy, chemical energy, electromagnetic energy, nuclear energy, or mass energy.Energy Transformations • Energy is always conserved • converted from one form into another • ‘initial state’ that transforms into a ‘final state’. , , ∆Ei ≡ Efinal i − Einitial i • Conservation of energy N ∆+ ∆ E2 + ... =∑∆Ei = 0E1 i=1System and Surroundings E∆ system + ∆ Esurroundings = 0Kinetic Energy • positive scalar quantity 12K = mv 2 • SI unit is defined to be a Joule 2 −21 J ≡ 1kg ⋅ m ⋅ sPRS Question 1 Compared to the amount of energy required to accelerate a car from rest to 10 miles per hour, the amount of energy required to accelerate the same car from 10 mph to 20 mph is 1) the same 2) twice as much 3) three times as much 4) four times as muchPRS Question 2 Consider two carts, of masses m and 2m, at reston an air track. If you push first one cart for 3 sand then the other for the same length of time,exerting equal force on each, the kinetic energyof the light cart is 1) larger than 2) equal to 3) smaller than the kinetic energy of the heavy car.Work Done by Constant Force Definition: Work done by a Constant Force The work done by the constant force acting on the body is the product of the component of the force in the direction displacement with the displacement, Wapplied = F ∆xxPRS Question 3 A ball is given an initial horizontal velocity and allowed to fall under the influence of gravity, as shown below. The work done by the force of gravity on the ball is: 1) positive 2) zero 3) negativePRS Question 4 A comet is speeding along a hyperbolic orbit toward the Sun. While the comet is moving away from the Sun, the work done by the Sun on the comet is: 1) positive 2) zero 3) negativeDot Product • The magnitude of the cross product GG ⋅=AB cos θAB • The dot product can be positive, zero, or negativeProjection and the Dot Product Two types of projections GG GG ⋅ =A(cos θ)B AB (cos θ)AB ⋅=ABProperties GG GG ⋅=⋅AB B A GG GG ⋅= c(AB)cAB ⋅ G GG G G GG (AB) C A C B C + ⋅= ⋅+⋅Unit Vectors and the Dot Product • Unit vectors ˆˆ ˆi i ˆ ii =|||| cos(0) =1 ⋅ˆˆ ˆˆ⋅k k jj =⋅=1 ˆˆ ˆ ˆ ij =|||j|cos(π/2) = 0 ⋅ i ˆ⋅=⋅ = 0ik ˆ ˆˆj kVector Components of Dot Product G i +A = A A A k ˆ ˆ j+ ˆ x y z G B =Bi + B B k ˆ ˆ j+ ˆ x yz GG ⋅=AB AB ABAB + +xx yy zzWork and the Dot Product GF ˆ ˆ j= FF i + xy ∆=∆x G x ˆi G ∆ Fx G ˆi ˆj ˆi FF xFW ( ) (∆) ∆+=⋅ = ⋅ = x y x xPRS Question 5 Consider two blocks stacked on a table. Someone pulls the bottom block to the right with a rope in such a way that both bocksaccelerate to the right but no slipping occurs at the interface between the top and bottomblocks. Friction at the interface between the two blocks does 1. Positive work on the top block. 2. No work on the top block. 3. Negative work on the top block.PRS Question 6 When a person walks, the force of friction between the floor and the person's feet accelerates the person forward. The floor does 1. Positive work on the person. 2. Negative work on the person. 3. No work on the person.Work-Kinetic Energy Theorem Wfi =∆= ∆= x(, F x ma x ma v t +1 at 2), x x xi x2 xf ,= m ,(v t + ( xf ,2t xi t (v , − vxi ) 1(v , − vxi ))t2 12 −= mvxf,1 mvxi 2 ,2 2 =∆K Kf =K Wfi +i ,Work-Kinetic Energy Theorem • one dimension xf xf dv vx,f xW =∫x0F dx =∫x0m dtdx =∫xx0fmdxdv =∫ mv dv x x x xdt vx,0 ⎛1 21 2mvx,f −2mvx,02 =∆ KW =∫vx,fmv dv =∫vx,fd ⎝⎜2mv 2 ⎞= 1 x x x ⎟vx,0 vx,0 ⎠Work done by Non-Constant Force • Infinitesimal work is a scalar ∆Wi = ()∆xiFx i • add up these scalar quantities to get thetotal work iN iN= = WN =∑∆Wi =∑()∆F xix i i1 i1= = • limit as N →∞ and ∆xi→ 0 iN xxf = = W lim∑()∆ ∫ x = F xi = FdxxiN→∞ x 0i1 xx0∆→= = iPRS Question 7 A particle starts from rest at x = 0 and moves to x = L under the action of a variable force F(x), which is shown in the figure. What is the particle's kinetic energy atx=L/2 and at x=L? 1. Fmax L/2, FmaxL 2. Fmax L/4, 0 3. Fmax L, 0 4. Fmax L/4, Fmax L/2 5. Fmax L/2, Fmax L/4Work Done Along an Arbitrary Path GG∆Wi = F ⋅ ∆ ri i =iN GG G W = lim ∑F ⋅∆ri =∫rr0f F r G ⋅ diNG→∞ = ∆→0i1riLine Integrals G ˆ• force vector F = Fˆi + Fˆj+ F kx y z G • line element dr = dxˆi + dy dz kˆj+ ˆ • total work GG GG GG GG rrf rrf rrf rrf = = = =GG ⋅ d ∫ y ∫ zW =∫ Fr =∫ Fdx + F dy + Fdz x GG GG GG GG rr0 rr0 rr0 rr0 = = = =Power • The average power of the applied force is defined to be the rate of doing work P = ,ave =∆W Fapplied , x ∆x =Fapplied , xvx ave ∆t ∆t • SI units of power are called Watts ⎣ ⋅−1 ⎦⎣ ⋅2 −3 ⎤≡ ⎦[1W ]≡⎡1 J s ⎤⎡1kg m ⋅sInstantaneous Power • limit of the average power ∆W ∆x v P =lim ∆t = Fapplied , x lim = Fapplied ,x x ∆→0 ∆→0 ∆tt tClass Problem 1 A person pushes a cup of mass m for a time t along a horizontal table with a force of magnitude F at an angle θ with respect to the horizontal for a distance d . The coefficient of friction between the table and the cup is µk. The cup was initially at rest.Class Problem 1 (con’t) a) Calculate the work done by the pushing force. b) Calculate the work done by the friction force. c) What is the average power of the pushing force? d) What is the average power of the kinetic friction force?MIT 8.01T Physics I Experiment …
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