#### Transcript Work-Energy Theorem

Chapter 6 Work, Energy, and Power 6.1 Work: The Scientific Definition - Many different forms of energy - Energy can be defined as the ability to do work, in some circumstances not all energy is available to do work. Work done on a system product of the component of the force in the direction of motions times the distance. F is the net force. A person holding a briefcase, does no work on it, since there is no motion. No work is done on the briefcase, and no energy is transferred it, when it is carried horizontally at a constant speed. Energy is transferred to the briefcase. Energy is transferred out the briefcase and into the electric generator. Here the work done on the briefcase by the generator is negative, since F and d are in opposite direction. F .d<0 Example Suppose the player in your game must push a secret stone wall at least 2m to gain access to a bonus level. If the player’s character can push with a horizontal force of 1500N and kinetic frictional force of 50N is working against him, how much work total will it take to move the wall out of the way? Solution • W = FDX = (1500-50)*2 = 2900 J Example Suppose you are coding a top-down game where the object being moved is free to move in both the x and y directions. A net force of 2000N@600 is applied to the object, but due to constraints, the object moves 3m@300. How much work is done to move the object? Solution • Use Dot product of two vectors • W = F.d = (2000cos 30)*3 = 5196 J F 300 Dx 6.2 Kinetic energy and the work-energy theorem Net W = (Net F) d - The package is accerlated from v0 to v by a net force F - Net F = ma, a=constant - v2 = v02 + 2ad Net W = ½ m (v2 - v02) = DKE Work-Energy Theorem the net work on a system = the changes in the kinetic energy Example Suppose you are programming a baseball game, and the shortstop need to throw the ball to the first baseman. It takes some work for him to get the ball moving . If he can produce a force of 400N for a distance of 0.75m (from behind his head to the point where he lets go of the ball), and the baseball has a mass of 145g, what speed should the ball have when he lets go? Solution • W = DKE • 400*0.75 = 0.5*0.145*(vf2 - 0) • vf = 64.3 m/s 6.3 Gravitational Potential Energy A mass is lifted up at constant speed W = Fd = mgh Define change in gravitational PE PEg(h)-PEg(0)= DPEg = mgh (Gravitational Potential energy, PEg) - if we release the mass, gravity will do an amount of work mgh on it Drop the mass, by the Work-energy Theorem Increasing the mass KE conversion of PE to KE PEg = 0 The change of gravitational PE (DPEg = mgh) between point A and B is independent of path. Gravity is a conservative force (守恒力) Example (a) v0 =0, mgd = DKE = ½ mvf2 vf = 19.8 m/s (b) v0 = 5m/s vf = 20.4 m/s - Mass cancel when friction is neglible - Speed depend on the initial position and final postion NOT on the PATH taken or mass (when friction is neglible) - Final speed 20.4 - 19.8=0.6 m/s << 5m/s Example Suppose you are programming an Indiana Jones game and you get to the part where he jumps into the mining cart and rides the track up and down a series of gills. If the cart is at a height of 100m when Indy jumps in, and together he and the cart weigh 136.18 Kg, how fast should they be going when they reach the bottom of the first hill (ground level)? Solution • ½ mvi2 + mgyi = ½ mvf2 + mgyf • 0 + 136.18*9.8*100 = 0.5*136.18*vf2 + 0 • vf2 = 44.27 m/s 100 m 6.4 Conservative Forces and Potential Energy 6.4 Conservative Forces and Potential Energy - Define the potential energy of a spring, - PEs = ½ kx2 (a general expression, such as atomic vibration) - It represents the work done on the spring and the energy stored in it as a result of strectching it a distance of x. 6.4 Conservative Forces and Potential Energy Work is done to deform the guitar string PE KE PE as the string oscillate back and forth A small fraction is dissipated as sound energy A nonconservative process 6.4 Conservative Forces and Potential Energy The net work done by all forces acting on a system equals to DKE, Net W = DKE (Work-energy Theorem ) If only conservative force involved Wc = DKE conservative force (gravity or a spring force) does work system loses PE Wc = mgh = - DPEg DKE + DPE = 0 KEf – KEi + PEf – PEi = 0 PEi + KEi= PEf + KEf Conservation of Mechanical Energy (PE+KE) h mg Example 6.7 A 100g toy car is propelled by a compressed spring, as shown in the figure. It follows a track that rises 0.18m above the starting point. The spring si compressed 4 cm and has a force constant of 250 N/m. Assuming no friction, find (a) how fast the car isi going before it starts up the slope and (b) how fast it is going at the tip of the slope, PEi + KEi= PEf + KEf 1/2mvi2+mghi+1/2kxi2 = 1/2mvf2+mghf+1/2kxf2 (a) 1/2kxi2 = 1/2mvf2 vf = (250/0.1)0.5 *0.04 = 2.0 m/s (b) 1/2kxi2 = 1/2mvf2+mghf vf = [(250/0.1) *0.042 – 2*9.8*0.18]0.5 = 0.69 m/s 6.5 Nonconservative Forces : Open systems - Frictional force, most of the energy goes into thermal energy - Open system energy may enter or leave it - Taken into account of both conservative and non-conservative forces Net W = Wc + Wnc = DKE -DPE + Wnc = DKE or Wnc = DPE + DKE (change in Mechanical Energy) PEi + KEi + Wnc = PEf + KEf Path taken by the frictional force is longer More energy lost Conservative Forces : Close systems Wnc = 0 6.5 Nonconservative Forces : Open systems 6.5 Nonconservative Forces : Open systems PEi + KEi + Wnc = PEf + KEf Work done by frictional force 6.6 Conservation of Energy • PEi + KEi + Wnc + OEi = PEf + KEf • OE = other energy fuel, electrical energy, chemical fuels, solar energy 6.7 Power P = W/t 1 hp = 746 Watt Example 6.9 What is the power output of mechanical energy for a 60 Kg woman who runs up a 3.0 m high flight of stairs in 2.5s, starting from rest but having a final speed of 2.0 m/s ? Solution W = DKE + DPE, if we set h=0, then both KE and PE are at the bottom. Thus W = KE + PEg = ½ mv2 + mgh P = W/t = [(0.5*60*2.02 + 60*9.8*3.0)]/2.5 = 754 W 6.8 Work, Energy, and Power in Humans; Efficiency Eff = Wout/Ein or Eout/Ein Example 6.11 Find the efficiency of the person in the figure who metabolized 8.0 kcal of food energy while lifting 120 kg to a height of 2.3 m above its starting point Solution The work output goes into increasing the gravitational PE of the weights, thus Wout = mgh Eff = Wout/Ein = 120*9.8*2.3/(8.0*4186) = 8.33% Table 6.3 Efficiency of he human body and mechanical devices Table 6.4 Energy and oxygen consumption rates (power)