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Work and Energy Physics Chapter 5 5.1 Work Scientific definition of work: product of force and displacement Example: pushing a shopping cart Equation: W = F d Explain the only time work is done. – Components of force is parallel to displacement Why is no work done by the student who is carrying a bucket of water? – Upward force exerted, moving horizontally 5.1 Work Net Work Done by a Constant Net Force W = Fnet d (cos θ) – Units: N (m) = Joules (J) 5.1 Work Example 1: How much work is done on a vacuum cleaner pulled 3.0 m by a force of 50.0 N at an angle of 30.0o above the horizontal? W = Fd (cos θ) W = 50.0 N (3.0 m) (cos 30o) W = 130 J 5.1 Work Example 2: A 20.0 kg suitcase is raised 3.0 m above a platform by a conveyor belt. How much work is done on the suitcase? Fg = mg Fg = 20 kg (9.81 m/s2) Fg = 196.2 N W = Fd W = 196 N (3.0 m) W = 588 J 5.2 Kinetic Energy Kinetic energy: energy associated with an object in motion Kinetic Energy KE = ½ mv2 – What type of quantity is kinetic energy? – Scalar (no direction = can’t be negative) Units: Joule 5.2 Kinetic Energy Kinetic energy depends upon what 2 things? – Speed (has the greater impact on KE) – Mass Example: bowling ball and volleyball – Traveling at same speed, bowling ball has more KE because more mass – Traveling at different speeds, volleyball may have more KE if moving a lot faster 5.2 Kinetic Energy Example 3: A 6.0 kg cat runs after a mouse at 10.0 m/s. What is the cat’s kinetic energy? KE = ½ mv2 KE = ½ (6.0 kg)(10.0 m/s)2 KE = 300 J 5.2 Kinetic Energy Example 4: A 7.00 kg bowling ball moves at 3.00 m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g table-tennis ball move in order to have the same kinetic energy as the bowling ball? Is this speed reasonable for a table-tennis ball? Bowling ball KE = ½ mv2 KE = ½ (7.0 kg)(3.0 m/s)2 KE = 31.5 J 5.2 Kinetic Energy Table Tennis Ball KE = ½ mv2 31.5 J = ½ (0.00245 kg)(v2) 31.5 J = 0.001225 (v2) 25714.29 = v2 v = 160 m/s ~345 mi/h 5.2 Work-Kinetic Energy Work-kinetic energy theorem: net work done by a net force acting on an object is equal to the change in KE Work-Kinetic Energy Theorem Wnet = ΔKE Wnet = KEf – KEi Wnet = ½mvf2 – ½mvi2 5.2 Work-Kinetic Energy Explain how we can tell if the speed of the object changes with the net work done. – Net work done is positive if speed increases This theorem allows us to think of kinetic energy as: – The work an object can do as it comes to rest, or amount of energy stored in the object 5.2 Work-Kinetic Energy Example: – Moving hammer has KE and can do work on the nail. – The KE is transferred to work on the nail – The hammer can only do as much work as it has KE 500 J of KE is converted to 500 J of work 5.2 Work-Kinetic Energy Example 5: A 1200 kg car is accelerated from 15.0 m/s to 25.0 m/s in a distance of 50.0 m. What net force acted upon the car? Wnet = ½mvf2 – ½mvi2 W = ½(1200 kg)(25.0 m/s)2 – ½ (1200 kg)(15.0 m/s)2 W = 3.75 x 105 J – 1.35 x 105 J W = 2.4 x 105 J W = Fd 2.4 x 105 J = F(50.0 m) F = 4.8 x 103 N 5.2 Work-Kinetic Energy Example 6: On a frozen pond, a person kicks a 10.0 kg sled, giving it an initial speed of 2.2 m/s. How far does the sled move if the coefficient of kinetic friction between the sled and the ice is 0.10? m = 10.0 kg vf = 0 m/s vi = 2.2 m/s µk = 0.10 Fnet = µk (m)(g) Fnet = 0.10 (10.0 kg)(9.81 m/s2) Fnet = -9.81 N 5.2 Work-Kinetic Energy W = ½mvf2 – ½mvi2 W = 0 J - ½ (10.0 kg)(2.2 m/s)2 W = -24.2 J Wnet = Fnet (d) -24.2 J = -9.81 N (d) d = 2.47 m 5.2 Potential Energy Potential energy: the amount of energy stored in an object What two things does the potential energy depend upon? – Properties of the object (ex: mass) – Interaction with environment (ex: gravity) 5.2 Potential Energy Gravitational PE: PE associated with an object due to the position of object relative to Earth or some gravitational source Gravitational Potential Energy PEg = mgh 5.2 Potential Energy Explain how gravitational PE must be measured. – Relative to a zero level – Example: volleyball A B C 5.2 Potential Energy Elastic PE: PE in a stretched/compressed elastic object Relaxed length: length when no external forces are acting on the spring Elastic Potential Energy PEelastic= ½ k x2 k = spring constant x = how far stretched/compressed 5.2 Potential Energy Spring constant: how resistant a spring is to being compressed or stretched – Small constant: flexible spring – Large constant: stiff spring 5.2 Potential Energy Example 7: A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river from a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling? PEtotal PEtotal = PEel + PEg PEtotal = ½ kx2 + mgh = ½(71.8 N/m)(29.0 m)2 + 70.0 kg(9.81)(6.0 m) PEtotal = 3.02 x 104 J + 4.1 x 103 J PEtotal = 3.43 x 104 J 5.2 Potential Energy Example 8: When a 2.0 kg mass is attached to a vertical spring, the spring is stretched 10.0 cm such that the mass is 50.0 cm above the table. What is the total potential energy associated with this mass relative to the table, if the spring constant is 15.0 N/m? PEtotal = PEel + PEg PEtotal = ½ kx2 + mgh PEtotal = ½(15.0 N/m)(0.10 m)2 + 2.0 (9.81)(0.50 m) PEtotal = 0.075 J + 9.81 J PEtotal = 9.89 J 5.3 Conservation of Energy Mechanical energy: sum of kinetic energy and all forms of potential energy Describe all forms of energy in a pendulum clock. – KE when pendulum is swinging – PEg when at different heights – PEelastic with springs, gears, etc. 5.3 Conservation of Energy Equation: ME = KE + PEg +PEelastic Nonmechanical energy: energy like nuclear, chemical, internal, and electrical 5.3 Conservation of Energy Conservation of Mechanical Energy MEi = MEf KEi + PEg,i + PEel,i = KEf +PEg,f +PEel,f ½mvi2 + mghi + ½kxi2 = ½mvf2 + mghf + ½kxf2 5.3 Conservation of Energy Example 9: Starting from rest, a child goes down a frictionless slide from an initial height of 3.00 m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0 kg. MEi = MEf KEi + PEgi = KEf + PEgf PEgi = KEf mghi = ½ mvf2 25.0 kg(9.81)(3.0 m) = ½ (25 kg)(vf2) 736 J = 12.5 (vf2) vf = 7.67 m/s 5.3 Conservation of Energy Example 10: A hurdler runs with a speed of 11.7 m/s. She jumps over the hurdle with a speed of 1.18 m/s. How high is she off the ground when going over the hurdle? (mass = 48.2 kg) MEi = MEf KEi + PEgi = KEf + PEgf KEi = KEf +PEgf ½ mvi2 = ½ mvf2 + mghf ½ (48.2)(11.7)2 = ½ (48.2)(1.18)2 + 48.2(9.81)(hf) 3299 J = 33.56 J + 472.84 (hf) 3265.4 J = 472.84(hf) hf = 6.9 m 5.3 Conservation of Energy When is mechanical energy not conserved? Total energy? – Mechanical: if friction is present – Total energy: always conserved 5.4 Power Power: rate at which energy is transferred; or the rate at which work is done Power P=W P = Fv Δt Units: Watts 5.4 Power Explain what makes one machine different from another. – All machines will do the same work – But in different time intervals Example: Horsepower in engines – To go from 0 to 60, more horsepower faster time – Lower HP, still get to 60 but in more time 5.4 Power Example 11: A 193 kg curtain needs to be raised 7.5 m, at constant speed, in as close to 5.0 s as possible. The power rating for three motors are listed at 1.0 kW, 3.5 kW, and 5.5 kW. Which motor is best for the job? P=W P = mgd Δt Δt P = 193 kg(9.81)(7.5 m) 5.0 s P = 2839 W = 2.8 kW 5.4 Power Example 12: Two horses pull a cart. Each exerts a force of 250.0 N at a speed of 2.0 m/s for 10.0 minutes. Calculate the power delivered by the horses. How much work is done by the two horses? P = Fv P = 250 N (2.0 m/s) P = 500 W x 2 horses = 1000 W P = W/ Δt 1000 W = W/600 s W = 6.0 x 105 J