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Sect. 7-4: Kinetic Energy; Work-Energy Principle

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## Sect. 7-4: Kinetic Energy; Work-Energy Principle

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**Energy: Traditionally defined as the ability to do work. We**now know that not all forces are able to do work; however, we are dealing in these chapters with mechanical energy, which does follow this definition. • Kinetic Energy The energy of motion “Kinetic” Greek word for motion An object in motion has the ability to do work.**Consider an object moving in straight line. Starts at speed**v1. Due to the presence of a net force Fnet, it accelerates (uniformly) to speed v2, over distance d. Newton’s 2nd Law: Fnet= ma (1) 1d motion, constant a (v2)2 = (v1)2 + 2ad a = [(v2)2 - (v1)2]/(2d) (2) Work done: Wnet = Fnet d (3) Combine (1), (2), (3):**Fnet= ma (1)**• a = [(v2)2 - (v1)2]/(2d) (2) • Wnet = Fnet d (3) Combine (1), (2), (3): Wnet = mad = md [(v2)2 - (v1)2]/(2d) or Wnet = (½)m(v2)2 – (½)m(v1)2**Summary:The net work done by a constant force in**accelerating an object of mass m from v1to v2is: DEFINITION:Kinetic Energy (KE) (for translational motion; Kinetic = “motion”) (units are Joules, J) • We’ve shown:The WORK-ENERGY PRINCIPLE Wnet = K( = “change in”) We’ve shown this for a 1d constant force. However, it is valid in general!**The net work on an object = The change in K.**Wnet = K The Work-Energy Principle Note!:Wnet = work done by the net (total) force. Wnetis a scalar & can be positive or negative (because K can be both + & -).If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases. The SI Units are Joulesfor both work & kinetic energy.**The Work-Energy Principle**Wnet = K NOTE! This isNewton’s 2nd Lawin Work & Energy Language!**A moving hammer can do work on a nail!**For the hammer: Wh = Kh = -Fd = 0 – (½)mh(vh)2 For the nail: Wn = Kn = Fd = (½)mn(vn)2 - 0**Example 7-7: Kinetic energy & work done on a baseball**A baseball, mass m = 145 g(0.145 kg) is thrown so that it acquires a speed v = 25 m/s. a. What is its kinetic energy? b. What was the net work done on the ball to make it reach this speed, if it started from rest?**Example 7-8: Work on a car to increase its kinetic energy**Calculate the net work required to accelerate a car, mass m = 1000-kg, from v1 = 20 m/s to v2 = 30 m/s.**Conceptual Example 7-9: Work to stop a car**A car traveling at speedv1 = 60 km/hcan brake to a stop within a distanced = 20 m. If the car is going twice as fast,120 km/h, what is its stopping distance? Assume that the maximum braking force is approximately independent of speed.**Wnet = Fd cos (180º) = -Fd(from the definition of work)**Wnet = K = (½)m(v2)2 – (½)m(v1)2(Work-Energy Principle) but, (v2)2 = 0 (the car has stopped) so-Fd = K= 0 - (½)m(v1)2 or d (v1)2 So the stopping distance is proportional to the square of the initial speed! If the initial speed is doubled, the stopping distance quadruples! Note:K (½)mv2 0Must be positive, since m & v2 are always positive (real v).**Example 7-10: A compressed spring**A horizontal spring has spring constant k = 360 N/m. Ignore friction. a. Calculate the work required to compress it from its relaxed length (x = 0) to x = 11.0cm. b. A1.85-kg block is put against the spring. The spring is released. Calculate the block’s speed as it separates from the spring at x = 0. c. Repeat part b. but assume that the block is moving on a table & that some kind of constant drag force FD = 7.0 N(such as friction) is acting to slow it down.**Example**A block of mass m = 6 kg, is pulled from rest (v0 = 0) to the right by a constant horizontal force F = 12 N. After it has been pulled for Δx = 3 m, find it’s final speed v. Work-Kinetic Energy Theorem Wnet = K (½)[m(v)2 - m(v0)2] (1) If F = 12 N is the only horizontal force, then Wnet = FΔx (2) Combine (1) & (2): FΔx = (½)[m(v)2 - 0] Solve for v: (v)2 = [2Δx/m] (v) = [2Δx/m]½ = 3.5 m/s FN v0**Conceptual Example**A man wants to load a refrigerator onto a truck bed using a ramp of length L, as in the figure. He claims that less work would be required if the length L were increased. Is he correct?**A man wants to load a refrigerator**onto a truck bed using a ramp of length L. He claims that less work would be required if the length L were increased. Is he correct? NO!For simplicity, assume that it is wheeled up the on a dolly atconstant speed. So the kinetic energy change from the ground to the truck isK = 0. The total work done on the refrigerator is Wnet = Wman + Wgravity + Wnormal The normal forceFNon the refrigerator from the ramp is at90º to the horizontal displacement & does no work on the refrigerator (Wnormal = 0).Since K = 0, by the work energy principle the total work done on the refrigerator is Wnet = 0. = Wman + Wgravity. So, the work done by the man isWman = - Wgravity.The work done by gravity isWgravity = - mgh[angle between mg & h is 180º & cos(180º) = -1]. So, Wman = mgh No matter what he does he still must do the SAME amount of work (assuming height h = constant!)