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

1 / 17

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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

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

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

### 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 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!

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 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 isK = 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!)