Sect 7 4 Kinetic Energy WorkEnergy Principle Energy

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

• 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 v 1. Due to the presence of a net force Fnet, it accelerates (uniformly) to speed v 2, over distance d. Newton’s 2 nd Law: Fnet= ma (1) 1 d motion, constant a (v 2)2 = (v 1)2 + 2 ad a = [(v 2)2 - (v 1)2]/(2 d) (2) Work done: Wnet = Fnet d (3) Combine (1), (2), (3):

• Fnet= ma • a = [(v 2)2 - (v 1)2]/(2 d) • Wnet = Fnet d (1) (2) (3) Combine (1), (2), (3): Wnet = mad = md [(v 2)2 - (v 1)2]/(2 d) or Wnet = (½)m(v 2)2 – (½)m(v 1)2

• 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 Wnet = K ( = “change in”) We’ve shown this for a 1 d 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. Wnet 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.

The Work-Energy Principle Wnet = K NOTE! This is Newton’s 2 nd Law in Work & Energy Language!

Table from another textbook

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

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.

Wnet = Fd cos (180º) = -Fd (from the definition of work) Wnet = 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).

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 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 (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 Wnet = K (½)[m(v)2 - m(v 0)2] (1) If F = 12 N is the only horizontal force, then Wnet = FΔx Combine (1) & (2): FΔx Solve for v: (2) = (½)[m(v)2 - 0] (v)2 = [2Δx/m] (v) = [2Δx/m]½ = 3. 5 m/s FN v 0

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 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 Wnet = Wman + Wgravity + Wnormal The normal force FN on the refrigerator from the ramp is at 90º 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 is Wman = Wgravity. The work done by gravity is Wgravity = - 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!)