sect. 7-4: kinetic energy; work-energy principle
TRANSCRIPT
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 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 v1 to v2 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 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 Newton’s 2nd 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 v1 = 20 m/s to v2 = 30 m/s.
Conceptual Example 7-9: Work to stop a car
A 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 (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 0 Must 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 = 11.0 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.
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 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 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!)