Work and Energy

Energy is often defined as the ability to do work (make things move).

Energy can be grouped into two basic forms: kinetic energy and potential energy.

If a energy can cause things to move it must be related in some way to the fundamental forces of nature.

Kinetic energy is the innate energy an object has because of its motion, while potential energy is “stored energy” and is related to an object’s position.

Work and energy, which turn out to be equivalent, are scalar quantities.

The unit of work and energy is the Joule (J). The joule can be defined in terms other base units (mks system) as you will see later.

Forms of Energy

When one talks of forms of energy (energy sources) one is usually referring to forms of potential energy.

Examples: chemical energy from fossil fuels ( electromagnetic forces—chemical bonds), nuclear energy, energy from the sun (nuclear strong force). Hydro (gravitational potential energy).

These forms of potential energy arise from the energy stored in force fields (fundamental forces)

e.g. Lifting an object in a the earth’s gravitational field stores (increases) gravitational potential energy. One had to do work against the gravitational force to accomplish this. Storing water behind a dam creates gravitational potential energy. If there was no gravity, there would be no pressure created by the weight of the water.

Kinetic Energy Kinetic energy is the energy associated with an objects motion. It is defined by:

EK =12mυ 2 = kinetic energy (J)

= mass (Kg) = speed (m/s)

EKmυ

Example: A 4.00 Kg object has 75.6 J of kinetic energy. How fast is it moving ?

υ =2EKm

= 6.15 m/s

Solution:

Work Done The work done by a constant force is defined as the distance moved multiplied by the component of the force in the direction of change in displacement: The angle 𝛳, is the angle between the

direction of force and the direction of change in displacement.

In the SI system, the units of work are joules:

As long as this person does not lift or lower the bag of groceries, he is doing no work on it. The force he exerts has no component in the direction of motion.

Example 1: An applied force (F= 30.0N) acts at an angle (𝛳) of 20.00 on a 4.00 kg mass (M) as shown, causing the block to move to the right. There is friction between the ground and the block of 12.0 N.

a)  Draw the complete FBD of the mass b)   Calculate the work done by each individual force over

a distance (d) of 10.0 m

F!

Md

Solution Example 1:

a) F!

M

FN

Fg

f

b) W= Fdcos𝛳 WFg= WFN = 0 J , since 𝛳= 900

WF= (30)(10)cos(200)= 282 J

Wf= (12)(10)cos(1800) = -120 J

Work done by forces can be negative or positive values. If the (component) force is in the direction of the change of displacement it is a positive value, and if the (component) force is in the opposite direction it is a negative value.

The work done by the net force (net work done) is equal to the sum of the work dones by all the individual forces. This can be tested by looking at the previous example-1 (work done over 10.0 m) .

The net force on the object was:

FNET= Fcos𝛳 –f= (30)cos20-12= 16.19 N

WFNET=FNETd= (16.19)(10)=161.9= 162 J

WFNET=WF + Wf + WFN + WFg= 281.91 + (-120) + 0 + 0 = 162 J

Comparing to the sum of all the work dones:

The work done is also defined as the area under the curve of a force (in direction of displacement) versus displacement (distance). Force

distance

F

d

Work= Area = FdFor a constant force

The work can be approximated by dividing the distance up into small pieces, finding the work done during each, and adding them up. As the pieces become very narrow, the work done is the area under the force vs. distance curve.

For a force that varies

Usually for varying forces problems constantly varying forces (straight line) are more common (see below). The calculation of the area of a triangle is straight forward.

Force

distance

F

d

Work = Area of triangle = 0.5Fd

Example 2: A constantly increasing force (F) acts on a 5.00 kg block over a distance of 15.0 m according to the graph below. What is the work done by the force F over this distance ?

MFF(N)

d(m)

4

8

5 10 15

(15,8)

Example 2- solution:

Split the area under the curve into two areas (rectangle and triangle)

F(N)

d(m)

4

8

5 10 15

(15,8)

A1

A2

Work = A1+A2 = (0.5)(15)(4) + (4)(15) = 90 J

13

Kinetic Energy and Work Done A car is at rest - its kinetic energy is 0 because ½mv2 = 0. A constant net force is then exerted on the car - work done on the car is: W = Fd but we also know that F = ma (Newton’s 2nd Law) Substitute F = ma for F in W = Fd then we have W = mad The car was originally at rest so its final velocity is given by: v2 = vo2 + 2ad but vo = 0 So v2 = 2ad or ad = ½v2 Substitute ad = ½v2 in W = mad and we have W = ½mv2. But not all objects start at rest - thus we can say that that Wnet = Ekf - Eko = ΔEk The change in kinetic energy of an object is equal to the net work done on it. This is known as the work-energy theorem. Also one can see that kinetic energy must share the same units as work done (Fd).

The Work-Energy Theorem

WNet = ΔEK

The net work done on an object (work done by net force) is equal to the object’s change in kinetic energy.

FNetd =12mυ f

2 −12mυi

2Putting in the details:

The work-energy theorem is basically a statement of energy conservation.

•  If the net work is positive, the kinetic energy increases. •  If the net work is negative, the kinetic energy decreases.

Example 3: An applied force (F= 30.0N) acts at an angle (𝛳) of 20.00 on a 4.00 kg mass (M) as shown, causing the block to move to the right. There is friction between the ground and the block of 12.0 N. If the block had an initial velocity of 3.00 m/s, using the work-energy theorem, determine its speed at 10.0 m.

F!

Md

Example 3-solution:

WNet = ΔEK WNet =12mυ f

2 −12mυi

2

161.91 = 12

(4)υ f2 −

12

(4)(3)2 WNet same as in example 1.

= 9.48 m/s υ f

Example 4 : A 2.50 gram bullet is fired from a rifle at 750 m/s horizontally into a block of material. If it penetrated 15.0 cm what average force (magnitude) stopped the bullet? Use the work-energy theorem. Start with a diagram.

Fd = 12mυ f

2 −12mυi

2

Solution:

F(0.15)= 0 – (0.5)(2.5x10-3)(750)2

F = - 4.69 kN Magintude of F = 4.69 kN

d!i

One expects a negative value for F in the calculation. The force in the opposite direction of d (change in displacement). If d is a taken to be positive F must be negative.

Potential Energy

An object can have potential energy by virtue of its surroundings.

Familiar examples of potential energy:

•  A wound-up spring

•  A stretched elastic band •  An object at some height above the ground

In raising a mass m to a height h, the work done by the external force is

We therefore define the gravitational potential energy:

Or EP =mghThe change in potential energy is then given by:

ΔEP = Epf −Epi =mghf −mghi

This potential energy can become kinetic energy if the object is dropped.

Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces).

If , where do we measure h from?

It turns out not to matter, as long as we are consistent about where we choose h = 0. Only changes in potential energy can be measured.

EP =mgh

Example 5: A 5.00 kg mass drops through a height of 15.0 m (hi) to a height of 10.0 m (hf) above the ground. a)  What was the object’s change in gravitational potential

energy? b)   What was the work done by the gravitational force?

hi

hf

d

ground

Example 5 solution:

a) ΔEP = Epf −Epi =mghf −mghi= (5)(9.8)(10)-(5)(9.8)(15)= -245 J

b)

= mgdcos𝛳 = (5)(9.8)(5)cos(0) = 245 J

One should notice that the change in gravitational potential energy is equal to negative work done by the gravitational force.

ΔEP = −WFg WFg = −ΔEPor

Potential energy can also be stored in a spring when it is compressed; the figure below shows potential energy yielding kinetic energy.

The force required to compress or stretch a spring is:

where k is called the spring constant, and needs to be measured for each spring.

The force increases as the spring is stretched or compressed further. We find that the potential energy of the compressed or stretched spring, measured from its equilibrium position, can be written:

Conservative and Non-conservative Forces

If friction is present, the work done depends not only on the starting and ending points, but also on the path taken. Friction is called a non-conservative force. Work done by conservative forces is path independent while work done by non-conservative forces is path dependent.

Potential energy can only be defined for conservative forces.

Therefore, we distinguish between the work done by conservative forces and the work done by non-conservative forces.

We find that the work done by non-conservative forces is equal to the total change in kinetic and potential energies:

If WNC= 0 , then ΔEK +ΔEP = 0

Mechanical Energy and Its Conservation

If there are no non-conservative forces (friction), the sum of the changes in the kinetic energy and in the potential energy is zero – the kinetic and potential energy changes are equal but opposite in sign.

This allows us to define the total mechanical energy (sum of kinetic and potential energies):

ETot = EP +EK ETot =U +K

If there are no non-conservative forces the total mechanical energy will remain constant (conserved). It should be noted that even if there is friction energy is still conserved (Law of Energy Conservation).

or (AP notation)

In the image on the left, the total mechanical energy is:

The energy buckets (right) show how the energy moves from all potential to all kinetic.

If there is no friction, the speed of a roller coaster will depend only on its height compared to its starting height.

A convenient way of stating energy mechanical energy conservation is the the following:

Ein +EPi +EKI = Eout +EPf +EKf Ein= energy input which increases total energy of system (+ value).

Eout= energy which leaves system. For example energy lost to friction (+value).

or

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

For the vast majority of problems Ein equals zero and if there is no friction Eout would also equal zero.

Example 6: A rollercoaster (1200 kg) initially at a height of 25.0 m (h1) had a speed of 5.00 m/s. a)  What is its total mechanical energy at this point ? b)   What speed would it have at a height of 10.0 m (h2) if

there was no energy lost to friction ?

h1

h2h3

Example 6 solution:

a) ETot = EP +EK =mgh1 +12mυ1

2 = (1200)(9.8)(25) + (0.5)(1200)(5)2

= 3.09x105 J

b) ETot = 3.09x105 =mgh2 +12mυ2

2

3.09x105=(1200)(9.8)(10) + (0.5)(1200)𝝊2

𝝊2= 17.9 m/s

Example 7: An 80.0 kg desperate student is skiing down a hill, fleeing from his physics teacher in order to avoid a challenging makeup test. He initially had a speed of 15.0 m/s (𝝊i). If he attains a final speed of i). If he attains a final speed of 30.0 m/s (𝝊f) at the base of the hill, and there was 10.0 kJ of energy lost to friction, what was the height (h) of the hill ?

h

υi

υ f

Example 7 solution:

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

0+ (80)(9.8)h + (0.5)(80)(15)2 = 1x104 + (80)(9.8)(0) + (0.5)(80)(30)2

h= 47 m

1.0m 0.50m

The bob of a pendulum has a mass of 3.0kg. This bob is pulled sideways so that it is 1.0m above the table top. a) What is the potential energy of the bob with respect to the table top? b) What is the potential energy of the bob with respect to the equilibrium position? c) What is the maximum velocity of the pendulum bob?

Example 8:

a) mass = 3.0kg g = 9.81m/s2 height = 1.0m Ep = mgh = 3.0kg x 9.81m/s2 x 1.0m = 29J.

Example 8 solution

c) Maximum velocity occurs at the equilibrium position, when all of the potential energy has been converted to kinetic energy. There is not friction so Eout is zero.

TME = Ek + Ep = 15J = ½mv2 + mgh = ½ (3.0kg) (v2) + 0J

Solving: v = 3.2m/s

To get this speed the mass of system is not actually necessary as m appears in every term of the relationship and can be cancelled. If there is friction one can not do this.

b) mass = 3.0kg g = 9.81m/s2 height = 0.5m

Ep = mgh = 3.0kg x 9.81m/s2 x 0.50m = 15J.

Example 9 ( A bit harder): A pendulum of 2.80 m in length (L) was released from rest an an angle of 370. If friction can be ignored what is the maximum speed attained by the pendulum ? Hints: 1) use the general energy conservation relationship as in the previous example 2) where would the the max speed occur? 3) To obtain the height above the base of it motion draw a horizontal line from it initial release point to create a right angle triangle.

!

"i

L

"f

Example 9 solution: Energy conservation states:

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

Ein and Eout are both zero. One needs to determine h, then one can solve for 𝝊f. The diagram below should make this more obvious.

L!Lcos!

h= 0h

L

According to the diagram: h = L- Lcos𝛳 = 0.564 m

There is enough information to solve for 𝝊f at the base of the pendulum’s motion using the above energy conservation relationship. One can divide each term by m to simplify the relationship.

(9.8)[2.8-2.8(cos(370))] = (0.5) υ f2

= 3.32 m/s υ f

Energy Conservation with Dissipative Processes

If there is a non-conservative force such as friction, where do the kinetic and potential energies go?

They become heat; the actual temperature rise of the materials involved can be calculated.

Energy is always conserved if one can keep track of where all the energy goes.

Summary of useful relationships for problem solving:

EK =12mυ 2 EP =mgh

WNet = ΔEK ΔEK +ΔEP = 0

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

The relationships below are not on any formula sheet

Another point

One often has the choice of a approaching a problem using vectors (e.g. forces and kinematics) or using a scalar approach (e.g. energy conservation). Using energy can simplify what could be a very difficult problem using forces (and kinematics). With the energy approach the problem only depends on the object’s beginning and end point and does not matter what happens in between (path taken by object). This can avoid some difficulty. For instance how would one attempt a swinging pendulum problem using forces(without using energy conservation)?

Free fall problems: 1. A 5.00kg object is at a height of 15.0m. What is its velocity just before it hits the ground if dropping from rest? (17.1 m/s) 2. From what height does a 25.0kg object free fall if its velocity just before it hits the ground is 42.4m/s? (91.7 m) Both problems may be solved using kinematics/dynamics and Energy methods – try both. This would also apply to 2-D projectile problems.

Example 10: A 12.0 kg object was fired of the edge of a cliff with an initial speed of 30.0 m/s (𝝊i)at an angle of 500 (𝛳) to the horizontal. What speed (𝝊f) i)at an angle of 500 (𝛳) to the horizontal. What speed (𝝊f) does it have 15.0 m (d) below the level it was fired at ? This could be done by standard projectile motion but it is much easier using energy conservation. Pick an convenient spot for the h=0.

!i

"

d!f

Example 10 solution:

I will use h=0 at its final location

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

Assuming there is no friction the above relationship is mass independent (divide each term by m)

(9.80)(15) + (0.5)(30.0)2 = (0.5) υ f2

υ f = 34.6 m/s

For practice try doing this problem with the standard 2-D projectile analysis.

Example 11: A 4.00 kg block was accelerated from rest up an incline (𝛳=300) to a speed of 5.00 m/s over a distance of 3.00 m. How much work (energy) was required to accomplish this ? There is a coefficient of friction between the block an incline of 0.450. One can use the general energy conservation relationship for this.

!f

!i

"

d

Example 11 solution:

Ein +mghi +12mυi

2 = Eout +mghf +12mυ f

2

h=0 at initial position. Ein would be the energy required to move the block to its second position. Eout is the energy lost to friction (+value).

Wf= fdcos𝛳 = (15.277)(3)cos(180) = - 45.83 J

f= 𝜇mgcos𝛳 = 15.277 N mgcos𝛳 = 15.277 N

∴ Eout = 45.83 J

h= dsin𝛳 = 1.50 m

Now use the conservation relationship to solve for Eout

Ein = 45.83 + (4.00)(9.80)(1.50) + (0.5)(4.00)(5.00)2 = 155 J

Power Power is the rate at which work is done. It is in general the rate of change of energy

The difference between walking and running up these stairs is power – the change in gravitational potential energy is the same.

In the SI system, the units of power are watts:

P =WΔt

=EΔt

This formula on the PH 12 formula sheet actually represents average power.

This is an appropriate time to introduce another very common unit of energy it is the kilowatt hour (kWh). Examining the units one can see that it is a product of power and time unit. E= PΔt =(kW)(h)

One can easily convert 1 kWh to joules. Recall that a W is a J/s:

1 KWh =1000 Wh= 1000 Js•h =1000 J

s•3600s = 3.60×106 J = 3.60 MJ

Example 12: A 60.0 W light bulb was left on for a full 23 days in a garage. How much energy was used by the bulb over this time period? State answer in Joules and kilowatt hours.

Solution: E = PΔt = (60)[(23)(24)(3600)] = 1.19 x108 J

= 33.1 kWh

The average power can be written in terms of the force and the average velocity:

P = Fυ

The power at instant (P) in time is where is the instantaneous velocity:

P = Fυ

Average power is the energy or work over a finite time interval. Instantaneous power is the rate of change of energy at an instant in time.

υ

Note that power P in this case is the rate a which work is done by a force.

Efficiency of a Machine (Device)

Any real machine can never be 100% efficient. For a machine to be 100% efficient it means that all the input put energy goes into performing the given task ( e.g. a electric motor that lifts an object ). This is impossible as there will always be some heat produced in the process

The efficiency of a machine can be calculated (in %) by the following relationship:

efficiency = e = useful energy outtotal energy in

×100

One can divide (top and bottom) by time and write this ration in terms of power.

e = useful power outtotal power in

×100

The useful energy is the minimum energy required to perform the given task assuming no heat loss by the device. For instance, in order to lift an object, the minimum energy required would be the change in gravitational potential energy.

Example 13 : A 20.0 W (input electrical power to device) electric motor is used to lift a 5.00 kg mass through a height of 2.00 m in a time of 6.00 s. What is the efficiency of the motor ?

Example 13 solution:

e = useful energy outtotal energy in

×100

=mgΔhPt

=(5)(9.6)(2)(20)(6)

×100 = 82%

Example 14 (harder): A 20.0 W (input electrical energy) electric motor pulls a 4.00 kg mass (M) up a 300 incline with friction (𝜇= 0.25) at a constant speed. The motor is 80 % efficient. At what speed is the block moving up the incline ? The machine has to do work against the frictional force and gravity. The sum of these two forces would just equal the force of tension since in translational equilibrium. The work done by tension is equal to the magnitude of the work done by the other two forces. One can relate work to force and power by: P=F𝝊

!M

motor

"

Solution Example 14:

Draw a FBD. Since the forces add to zero one can state:

FT = mgsin𝛳 + 𝜇mgcos30 = 28.087 N

Pinput = 20 W ∴ Puseful = 16 W e =useful power out

total power in

FT𝝊=P 𝝊= P/FT = (16)/(28.087) = 0.57 m/s