AP Physics Chapter 7 Kinetic Energy and Work 1 AP Physics 2 Turn in Chapter 6 Homework, Worksheet & Lab Report Lecture Q&A Kinetic Energy, K Kinetic Energy: ability to do work due to motion 1 2 K mv 2 3 – – m: mass of object v: velocity of object Unit: [K] = [m] · [v]2 = kg · (m/s)2 = Joule = J 1 electron-volt = 1 eV = 1.60 10-19 J Scalar, no direction K0 Example: What is the kinetic energy associated with the Earth’s orbiting of the Sun? (Mass of Earth is 5.98 1024 kg, and radius of Earth orbiting around sun is 1.50 1011 m.) m = 5.98 1024 kg, T = 365 dys = 3.15 107 s r = 1.50 1011 m, K = ? 2 r v T 2 1.50 1011 m 3.15 107 s m 2.99 10 s 4 2 1 2 1 24 4 m 33 K mv 5.98 10 kg 2.99 10 2.67 10 J 2 2 s 4 Work, W Work: energy transferred to or from an object by means of a force acting on the object Work done by object A (to object B) 5 W > 0 if energy transferred (from A) to object B W < 0 if energy transferred from object B (to A) F d Work Done by A Constant Force W F d Fd cos Fx d x Fy d y Fz d z W: work done by force on the object F: constant force acting on object d: displacement of object with force acting on, x : angle between F and d 6 < 90o W > 0 > 90o W < 0 = 90o W = 0 W F d W: a scalar, no direction (but W can be negative) Valid for particle-like objects Work Done by Constant Forces W F d F 1 F 2 F 3 ... d F1 d F 2 d F 3 d ... W1 W2 W3 ... 7 Find the net force (vectors), then find the dot product of the net force and the displacement, or Find the work done by each individual force, then add up all the works done by each force. Practice: Pg160-8 A floating ice block is pushed through a displacement d = (15 m)i – (12 m) j along a straight embankment by rushing water, which exerts a force F = (210 N)i – (150N )j on the block. How much work does the force do on the block during the displacement? ˆ ˆ d 15mi 12m j, F 210N iˆ 150N ˆj, W ? W F d Fx dx Fy d y 210 N 15m 150 N 12m 3200 J 1800 J 5000 J 5.0 103 J 5.0kJ 8 Work-Kinetic Energy Theorem The total work done on an object is equal to the change in the kinetic energy of that object W K K f Ki Another look at work: W F d F vt F v t If F and d are in same direction: W Fd 9 Practice: A single force acts on a body that moves along a straight line. The diagram shows the velocity versus time for the body. Find the sign (plus or minus) of the work done by the force on the body in each of the intervals AB, BC, CD, and DE. v W = K AB: BC: 10 K: __ 0 __ + K __ > 0 WAB __ > 0 v __ = 0 K __ = 0 WBC __ = 0 B C + A - CD: K: __ + __ 0 K __ < 0 WCD __ < 0 DE: K: __ 0 __ + (though v: __ 0 __) - WDE __ > 0 D t E Practice: Pg160-13 Figure 7-29 shows three forces applied to a greased trunk that moves leftward by 3.00 m over a frictionless floor. The force magnitudes are F1 = 5.00 N, F2 = 9.00 N, and F3 = 3.00 N. During the displacement, (a) what is the net work done on the trunk by the three forces and (b) does the kinetic energy of the trunk increase or decrease? F1 5.00 N , 1 0o , F2 9.00 N , 2 120o , F3 3.00 N , 3 90o , d 3.00m, a ) W ? F2 60o F1 W Fd cos W1 5.00N 3.00m cos0o 15.0J W2 9.00N 3.00m cos120o 13.5J W3 3.00N 3.00m cos90o 0J W W1 W2 W3 15.0J 13.5J 0 1.5J b)K W 0 K increases 11 F3 W Fd cos Work (Wg) Done by Weight (Fg) Wg mgd cos m: mass g = 9.8 m/s2 d = displacement : angle between d and downward direction h 180o Wg mg h 12 mg h: change in object’s height ? Going up h > 0 Wg < 0 Weight doing negative work Going down h __ ______ work < 0 Wg ___ > 0 Weight doing positive Practice: Pg164-59 To push a 25.0 kg crate up a frictionless incline, angled at 25.0o to the horizontal, a worker exerts a force of 209 N, parallel to the incline. As the crate slides 1.50 m, a) How much work is done on the crate by the worker’s applied force, b) How much work is done on the crate by the weight of the crate, and c) How much work is done on the crate by the normal force exerted by the incline on the crate? d) What is the total, work done on the crate? m 25.0kg, 25.0o , F 209N , d 1.50m a) 0o ,Wapp ? Wapp Fd cos 209N 1.50m cos0o 314J d b)Wg ? s2 g W mg h mgd sin 25.0kg 9.8 m 1.50m sin 25.0o 155 J c)WN ? 90o WN 0 d ) W ? W Wapp Wg WN 314 J 155J 159 J 13 h Practice: Pg161-19 A cord is used to vertically lower an initially stationary block of mass M at a constant downward acceleration of g/4. when the block has been lowered a distance d, find a) the work done by the cord’s force on the block, b) the work done by the weight of the block, c) the kinetic energy of the block, and d) the speed of the block. 14 Solution: Pg 161-19 Let downward be positive direction, then F W T ma mg / 4 T W mg / 4 mg mg / 4 3mg / 4 a) 180o ,WT ? WT Td cos 3mg / 4 d cos180o 3mgd / 4 T b) 0o ,Wg ? Wg mgd cos 0o mgd + c) K f ? W K K f Ki W WT Wg 3/ 4mgd mgd mgd / 4 K f Ki mgd / 4 mgd / 4 d )v ? 15 2 mgd / 4 1 2 2K v K mv m m 2 gd 2 Work Done by a Variable Force xf W Fd W F ( x)dx 1-D xi xf W yf F ( x)dx F ( y )dy F ( z )dz xi yi or: Area under curve of Force-position graph 16 zf zi 3-D Example: Pg162-31 A 10 kg mass moves along an x axis. Its acceleration as a function of its position is shown in Fig. 7-37. What is the net work performed on the mass by the force causing the acceleration as the mass moves from x = 0 to x = 8.0 m? m 10kg, xi 0, x f 8m,W ? xf W F ( x)dx xi xf ma( x)dx xi xf m a ( x )dx xi m Area Under Curve m 1 10kg 8m 20 2 800 J s 2 17 Spring Force When a spring is compressed or stretched, it exerts a force on the object that compresses or stretches it. Force is opposite to change of length of spring Restoring force 18 F x F x Hooke’s Law Spring force is given by F k x or F kx F: Force exerts by spring, not by you k: Spring constant, property of spring, stiffness of spring (larger k, stiffer spring) x: Displacement of (end of) spring from equilibrium position, x -: F and x are in opposite direction Notice that this force is the force exerted by the spring, not the force you apply to compress or stretch the spring. 19 Ideal (Perfect) Spring 20 No mass Can always go back to original length when not compressed or stretched. Does not become warm/hot x is small compared to the entire length of spring Work Done by Spring (Force) Ws x Fdx x kx dx xf xf i i 1 2 1 kxi kx f 2 2 2 This is the work done by the spring, not the work done (by you) to compress or stretch the spring. If the spring is initially at equilibrium position, then xi = 0, 1 2 1 1 and 2 2 Ws kxi kx f kx f 0 2 2 2 F 21 Ws xf x - work ? x Xi = 0 F xf Example: A spring with a spring constant of 15 N/cm has a cage attached to one end. How much work does the spring force do on the cage if the spring is stretched from its relaxed length by 7.6 mm? How much additional work is done by the spring force if it is stretched by an additional 7.6 mm? equilibrium 7.6 mm k 15 7.6 mm N 100cm N 1500 cm m m a) xi 0, x f 7.6mm 7.6 103 m,Ws ? 2 1 N 1 2 1 3 2 1500 7.6 10 m 0.043J Ws kxi kx f 2 m 2 2 b) xi 7.6 103 m, x f 2 7.6mm 1.52 102 m,Ws ? 1 2 1 1 kxi kx f 2 k xi 2 x f 2 2 2 2 2 2 1 N 3 2 1500 7.6 10 m 1.52 10 m 0.13J 2 m Ws 22 Practice: The only force acting on a 2.0 kg body as it moves along the positive x axis has an x component Fx = -6x N, where x is in meters. The velocity of the body at x = 3.0 m is 8.0 m/s. a) What is the velocity of the body at x = 4.0 m? b) At what positive value of x will the body have a velocity of 5.0 m/s? 23 m 2.0kg, k 6N / m, xi 3.0m, vi 8.0m / s a) x f 4.0m, v f ? W Solution K xf xi Fdx 4.0 3.0 6x dx 3x 3.0 2 4.0 1 1 1 2W mv f 2 mvi 2 m v f 2 vi 2 W v f 2 vi 2 2 2 2 m vf 2W vi 2 m 2 21J m m 8.0 6.6 2.0kg s s 2 b)v f 5.0m / s, x f ? 2 2 1 1 m m 2 2 W K m v f vi 2kg 5.0 8.0 39 J 2 2 s s W 24 xf xi xf Fdx 3.0 6 x dx 3x 3.0 3x x xf 2 xf 27 39 27 W 4.7m 3 3 2 3.0 f 27 3x f 2 21J Power Power: the (time) rate at which work is done by a force, or the (time) rate at which energy is transferred (from one object to another) Average Power: W P – – 25 t W: work (scalar, no direction) t: time, duration, for that amount of work done Unit of Power: W P t J Watt W ; s 1 horsepower = 1 hp = 746 W Instantaneous Power d F x dx dW F F v P dt dt dt Fv cos or Fx vx Fy v y Fz vz F: force, vector v: velocity, vector : angle between F and v 26 Example: Pg165-73 An elevator cab has a mass of 4500 kg and can carry a maximum load of 1800 kg. If the cab is moving upward at full load of 3.80 m/s, what power is required of the force moving the cab to maintain that speed? M m1 m2 4500kg 1800kg 6300kg , m v 3.80 , P ? s m F Mg 6300kg 9.8 2 61700 N s 0 m P Fv cos 61700 N 3.80 2.34 105W s 27 Practice: Pg162-44 a) b) 28 At a certain instant, a particle experiences a force F = (4.0 N)i – (2.0 N)j + (9.0 N)k while having a velocity v = -(2.0 m/s)i + (4.0 m/s)k. What is the instantaneous rate at which the force does work on the particle? At some other time, the velocity consists of only a j component. If the force is unchanged, and the instantaneous power is –12 W, what is the velocity of the particle just then? Solution: Pg162-44 m m a) F 4.0 N iˆ 2.0 N ˆj 9.0 N kˆ, v 2.0 iˆ 4.0 kˆ, P ? s s P F v Fx vx Fy v y Fz vz m m 4.0 N 2.0 2.0 N 0 9.0 N 4.0 28W s s b) v = ? Let the velocity be v = vy j, then P F v Fx v x Fy v y Fz v z 2.0 N v y vy 29 P 12W m 6.0 2.0 N 2.0 N s So, v = (6.0m/s) j Practice: Pg162-45 A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min. The elevator’s counterweight has a mass of only 950 kg. so, the elevator motor must help pull the cab upward. What average power (in horsepower) is required of the force that motor exerts on the cab via the cable? 30 Solution: Pg162-45 me 1200kg , mc 950kg , h 54m, t 3min P? Three things doing work: Work done by gravity on elevator cab: We = -meg h = -1200kg 9.8m/s2 54 m = -6.35 105 J Work done by gravity on counterweight: Wc = -mcgh = -950kg 9.8m/s2 (-54 m) = 5.03 105 J Work done by motor: Wm = ? Constant velocity: W = 0 Wm + We + Wc = 0 So, Wm = -(We + Wc ) = - (-6.35 105 J + 5.03 105 J) = 1.32 105 J Then average power is 31 Wm 1.32 105 J min hp P 733W 0.98hp t 3.0 min 60s 746W Another approach The motor essentially needs to do work to raise the cab (1200 kg) 54 m and lower the counterweight (950kg) 54 m. So equivalently, the motor needs to do work to raise a mass of 250 kg (1200kg – 950kg) 54 m. The force needed is F mg 250kg 9.8 m 2450 N 2 s So the work done by the motor is W Fd 2450 N 54m 132300 J And the power is P 32 W 132300 J 735W 0.98hP t 180s Reference Frames Invariant quantities: quantities measured exactly the same at different inertial reference frames – Variant quantities: – force, mass, acceleration, time displacement, velocity Principle of Invariance: The laws of physics must have the same form in all inertial reference frames. 33