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

2. 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.

3. Consider an object moving in straight line. Starts at speed v 1. Due to the presence of a net force F net, it accelerates (uniformly) to speed v 2, over distance d. Newton’s 2 nd Law: F net = ma (1) 1d motion, constant a  (v 2 ) 2 = (v 1 ) 2 + 2ad  a = [(v 2 ) 2 - (v 1 ) 2 ]/(2d) (2) Work done: W net = F net d (3) Combine (1), (2), (3):

4. F net = ma (1) a = [(v 2 ) 2 - (v 1 ) 2 ]/(2d) (2) W net = F net d (3) Combine (1), (2), (3):  W net = mad = md [(v 2 ) 2 - (v 1 ) 2 ]/(2d) or W net = (½)m(v 2 ) 2 – (½)m(v 1 ) 2

5. Summary: The net work done by a constant force in accelerating an object of mass m from v 1 to v 2 is: DEFINITION: Kinetic Energy (KE) (for translational motion; Kinetic = “motion”) (units are Joules, J) We’ve shown: The WORK-ENERGY PRINCIPLE W net =  K (  = “change in”) We’ve shown this for a 1d constant force. However, it is valid in general!

6. The net work on an object = The change in K. W net =  K  The Work-Energy Principle Note!: W net = work done by the net (total) force. W net is 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 Joules for both work & kinetic energy.

7. The Work-Energy Principle W net =  K NOTE! This is Newton’s 2 nd Law in Work & Energy Language!

8. Table from another textbook

9. A moving hammer can do work on a nail! For the hammer: W h =  K h = -Fd = 0 – (½)m h (v h ) 2 For the nail: W n =  K n = Fd = (½)m n (v n ) 2 - 0

10. 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?

11. 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 v 1 = 20 m/s to v 2 = 30 m/s.

12. Conceptual Example 7-9: Work to stop a car A car traveling at speed v 1 = 60 km/h can brake to a stop within a distance d = 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.

13. W net = Fd cos (180º) = -Fd (from the definition of work) W net =  K = (½)m(v 2 ) 2 – (½)m(v 1 ) 2 (Work-Energy Principle) but, (v 2 ) 2 = 0 (the car has stopped) so -Fd =  K = 0 - (½)m(v 1 ) 2 or d  (v 1 ) 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)mv 2  0 Must be positive, since m & v 2 are always positive (real v).

14. 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 = 11.0 cm. b. A 1.85-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 F D = 7.0 N (such as friction) is acting to slow it down.

15. Example A block of mass m = 6 kg, is pulled from rest (v 0 = 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 W net =  K  (½)[m(v) 2 - m(v 0 ) 2 ] (1) If F = 12 N is the only horizontal force, then W net = 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 FNFN v0v0

16. 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?

17. 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 at constant speed. So the kinetic energy change from the ground to the truck is  K = 0. The total work done on the refrigerator is W net = W man + W gravity + W normal The normal force F N on the refrigerator from the ramp is at 90º to the horizontal displacement & does no work on the refrigerator (W normal = 0). Since  K = 0, by the work energy principle the total work done on the refrigerator is W net = 0. = W man + W gravity. So, the work done by the man is W man = - W gravity. The work done by gravity is W gravity = - mgh [angle between mg & h is 180º & cos(180º) = -1]. So, W man = mgh No matter what he does he still must do the SAME amount of work (assuming height h = constant!)