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 motionKinetic Greek word for motionAn 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.
Newtons 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 = ()m(v2)2 ()m(v1)2
Summary: The net work done by a constant force in accelerating an object of mass m from v1 to v2 is:
DEFINITION: Kinetic Energy (KE) (for translational motion; Kinetic = motion) (units are Joules, J)
Weve shown: The WORK-ENERGY PRINCIPLEWnet = K ( = change in)Weve 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.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 Newtons 2nd Law inWork & Energy Language!
Table from another textbook
A moving hammer can do work on a nail!
For the hammer:Wh = Kh = -Fd = 0 ()mh(vh)2
For the nail:Wn = Kn = Fd = ()mn(vn)2 - 0
Example 7-7: Kinetic energy & work done on a baseballA 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 carA car traveling at speed v1 = 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 (180) = -Fd (from the definition of work)Wnet = K = ()m(v2)2 ()m(v1)2 (Work-Energy Principle)but, (v2)2 = 0 (the car has stopped) so -Fd = K = 0 - ()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 ()mv2 0 Must be positive, since m & v2 are always positive (real v).
Example 7-10: A compressed springA 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 blocks 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.
ExampleA 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 x = 3 m,find its final speed v. Work-Kinetic Energy Theorem Wnet = K ()[m(v)2 - m(v0)2] (1)If F = 12 N is the only horizontal force,then Wnet = Fx (2) Combine (1) & (2): Fx = ()[m(v)2 - 0] Solve for v: (v)2 = [2x/m] (v) = [2x/m] = 3.5 m/s
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 90 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 180 & cos(180) = -1]. So, Wman = mgh No matter what he does he still must do the SAME amount of work (assuming height h = constant!)
Figure 7-18.Solution: For (a), use the work needed to compress a spring (already calculated). For (b) and (c), use the work-energy principle. W = 2.18 J. 1.54 m/s The drag force does -0.77 J of work; the speed is 1.23 m/sFigure 7.13: (Example 7.6) A block pulled to the right on a frictionless surface by a constant horizontal force.Figure 7.14: (Conceptual Example 7.7) A refrigerator attached to a frictionless, wheeled hand truck is moved up a ramp at constant speed.Figure 7.14: (Conceptual Example 7.7) A refrigerator attached to a frictionless, wheeled hand truck is moved up a ramp at constant speed.