Chapter 6 – Work and Energy

Midterm exams will be available next Thursday.

AssignmentAssignment 66Textbook (Giancoli, 6 th edition), Chapter 6:

Due on Thursday, November 5

1. On page 162 of Giancoli, problem 4.



4. A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1.00 m/s and then has the same kinetic energy as the boy. What werethe original speeds of man and boy?

Chapter 6

• Work Done by a Constant Force

• Kinetic Energy, and the Work-Energy Principle

• Potential Energy• Potential Energy

• Conservative and Nonconservative Forces

• Mechanical Energy and Its Conservation.

Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture

Kinetic Energy, and the Work Energy PrincipleKinetic Energy, and the Work Energy Principle

In mechanics, we can use a less precise definition of energy as:

“The ability to do work”

The energy of motion is called kinetic energy (or KE for short).


Work-energy principle :

“ The net work done on a object is equal to the chan ge in the object’s kinetic energy ”

1) If Wnet > 0, then ∆KE > 0 implying that there was an increase in the object’s velocity ( v2 > v1 ).

2) If Wnet < 0, then ∆KE < 0 implying that there was an decrease in the object’s velocity ( v2 < v1 ).


Problem 6-25 (textbook): A 285-kg load is lifted 22.0 m vertically with an acceleration a = 0.160g by a single cable. Determine

(a) the tension in the cable,

(b) the net work done on the load,

(c) the work done by the cable on the load,

(d) the work done by gravity on the load, and (d) the work done by gravity on the load, and

(e) the final speed of the load assuming it started from rest.


Problem 6-25:

(a)

From the free-body diagram for the load being lifted, write Newton’s 2nd law for the vertical direction, with up being positive.

( )( )T

2 3

T

0.160

1.16 1.16 285 kg 9.80 m s 3.24 10 N

F F m g m a m g

F m g

= − = = →

= = = ×

∑Fnet =

( )( )T

TFr

m gr


Problem 6-25:

(b)

The net work done on the load is found from the net force.

( ) ( )( )( )o 2

net net

3

cos 0 0.160 0.160 285 kg 9.80 m s 22.0 m

9.83 10 J

W F d mg d= = =

= ×

TFr

m gr


Problem 6-25:

(c)

The work done by the cable on the load is

( ) ( )( )( )o 2 4

cable T cos 0 1.160 1.16 285 kg 9.80 m s 22.0 m 7.13 10 JW F d mg d= = = = ×

TFr

m gr


Problem 6-25:

(d)

The work done by gravity on the load is

( ) ( )( )o 2 4

G cos180 285 kg 9.80 m s 22.0 m 6.14 10 JW mgd mgd= = − = − = − ×

TFr

m gr

Work Done by a Constant ForceWork Done by a Constant Force

Problem 6-25:

(e)

Use the work-energy theory to find the final speed, with an initial speed of 0.

( )

2 21 1n e t 2 1 2 12 2

3

2

2 1

2 9 .8 3 1 0 J20 8 .3 1 m s

2 8 5 k gn e t

W K E K E m v m v

Wv v

m

= − = − →

×= + = + =

TFr

m gr

2 8 5 k gm

Another form of energy “carried” by an object is called potential energy.

Potential energy is associated to forces that depend on the position or configuration of an object.

There are several forces that depend on either the configuration or position of an object

� therefore there are several different kinds of potential energies , eachassociated to a different force .

Potential EnergyPotential Energy

associated to a different force .

For example, we have seen that the gravitational force depends on the distancebetween two objects

� therefore, the energy of an object associated to the gravitational force (gravitational potential energy ) will consequently depend on the position of the object.

Let’s analyze the gravitational potential energy …………..

In raising a mass m to a height h (no net acceleration) , the work done by the external force is

Gravity also does work on the object of mass m:


(6-8)

Gravity also does work on the object of mass m:

Now, note that if at any instant of time you dropthe object , it can do work on something located underneath it. Let’s say it strikes a nail, moving it into the floor.

� it will change the nail’s velocity, therefore it will change the nail’skinetic energy.

How does it work?

As you drop the object, it will immediatelygain velocity due to the acceleration ofgravity.

Assuming its initial velocity to be zero, we have for its velocity when it strikes the nail (travels a distance h):


Therefore, using the work-energy principle, we have:

This results tells us that an object located at a height h has the ability to do an amount of work given by mgh

0 2gh

The previous result can be generalized to anyheight y such that we can define agravitational potential energy as

Given that h = y2 – y1 , it follows that:

which is exactly eq. 6.8 obtained two slides ago.


(6-9)

which is exactly eq. 6.8 obtained two slides ago.

We can rewrite the above equation as:

Note 1: The choice of y=0 is arbitrary � only changes in potential energy matter .

Note 2: Again, from two slides ago we can conclude that:

(6-10)

Important:

1) Potential energy is a property of a system as a whole .

For example, when you lift the object in the figure from y1 to y2, the work done is given bythe difference between PE2 and PE1.

The values of PE and PE depend on the object


The values of PE1 and PE2 depend on the objectand on the Earth (responsible for the force of gravity),therefore on the complete system : Earth + object .

This is NOT the case with kinetic energy whereONLY the properties of the object matters

2) The work done by the gravitational force on an object depends only on the height and not on the path taken by the object.

As mentioned before, there are other forms of potential energy. I will now discuss on one of them associated with elastic materials.

A spring has potential energy when compressed or stretched.

For example, the figure below shows the potential energy stored in a spring when it is compressed yielding kinetic energy when it is released.

Potential Potential Energy (Elastic force)Energy (Elastic force)

The force needed to compress or stretch a string is written as:

Where,

k = spring stiffness constant

Newton’s 3rd law says that the spring willexert a force FS in reaction to the forceF such that:

Potential Energy (Elastic force) Potential Energy (Elastic force)

SFP such that:

Textbooks usually write:

Eq. 6.11 (spring equation) is known as Hooke’s law

FS is the force the spring exerts opposite to its displacement in an attempt to restore its original length

(6-11)

Now that we know the force, we can define the potential energy of a spring compressed or stretched by a length x from an initial reference position.

The work done to change the spring’s natural length, let’s say stretch it, will be given by:

Where,

(We cannot use F directly to calculate the work

Potential Energy (Elastic force)Potential Energy (Elastic force)

(We cannot use FP directly to calculate the workdone on the string since FP = kx varies with the position x. Thus, we use the average force applied on the spring).

Thus

We can define the elastic potential energy as:

(6-12)


Problem 6-30 (textbook): A 1.60-m tall person lifts a 2.10-kg book from the ground so it is 2.20 m above the ground. What is the potential energy of the book relative to

(a) the ground

(b) and the top of the person’s head?

(c) How is the work done by the person related to the answers in parts (a) and (b)?


Problem 6-30:

(a) Relative to the ground, the PE is given by

(b) Relative to the top of the person’s head, the PE is given by

( ) ( )( )( )2

G book ground 2.10 kg 9.80 m s 2.20 m 45.3 JPE mg y y= − = =

( ) ( )( )( )22.10 kg 9.80 m s 0.60 m 12 JPE mg y y h= − = =

(c) The work done by the person in lifting the book from the ground to the final height is the same as the answer to part (a), 45.3 J.

In part (a), the PE is calculated relative to the starting location of the application of the force on the book. The work done by the person is not related to the answer to part (b).

( ) ( )( )( )2

G book head 2.10 kg 9.80 m s 0.60 m 12 JPE mg y y h= − = =


Problem 6-29 (textbook): A 1200-kg car rolling on a horizontal surface has speed v = 65 Km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?


Problem 6-29:

Assume that all of the kinetic energy of the car becomes PE of the compressed spring.

( ) ( )

( )

2

22 2 41 1

2 2 22

1m s1200 kg 65 km h

3.6 km h 8.1 10 N m

2.2 m

mvmv kx k

x= → = = = ×

( )

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