# Work and Energy - PowerPoint PPT Presentation

1 / 47

Work and Energy. Definitions. Work – the product of force and the component of displacement in the direction of the force. - work is a scalar quantity - Without motion there is no work. W = F ∙ d The unit of work is the newton meter, which is called a joule (J)

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

Work and Energy

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

## Work and Energy

### Definitions

• Work

– the product of force and the component of displacement in the direction of the force.

- work is a scalar quantity

- Without motion there is no work.

W = F ∙ d

The unit of work is the newton meter, which is called a joule (J)

(in honor of English Scientist James Prescott Joule)

### Example

• How much work is done on an object if a force of 30 Newtons [south] displaces the object 200 meters [south]?

• Solution:W = F∙d

= (30 N[S])(200 m [S])

= 6000 J

### Work (cont’d)

• Suppose force and displacement are not in the same direction.

• Work is defined to be the product of the force in the direction of the displacement and the displacement

F

W = (Fcosθ)∙d

d

θ

object

### example

• As Alex pulls his red wagon down the sidewalk, the handle of the wagon makes an angle of 60 degrees with the pavement. If Alex exerts a force of 100 Newtons along the direction of the handle, how much work is done when the displacement of the wagon is 20 meters along the ground?

## Power

### Definition

• Power – the rate at which work is done

Power is also a scalar quantity, and its unit is Joules per second (J/s), also known as a watt(W).

### Example

• If 300 J of work is performed on an object in 1.0 minute, what is the power expended on the object?

P = W / t

P = 300J / 60 sec

P = 5 W

### Example 2

• A 200 N force is applied to an object that moves in the direction of the force. If the object travels with a constant velocity of 10m/s, calculate the power expended on the object.

P = 200N (10 m/s)

P = 2000 W

### Hw

• Read pg. 80-83 Do # 1-25

## Energy

Mechanical Energy

### Kinetic Energy

• Work done on an object changes its Kinetic Energy.

• Therefore, W = KEf – KEi = ΔKE.

• The formula for Kinetic Energy is

KE = ½ mv2

### Example

• A 10 kg object subjected to a 20. N force moves across a horizontal, frictionless surface in the direction of the force. Before the force was applied, the speed of the object was 2.0m/s. When the force is removed the object is traveling at 6.0 m/s. Calculate the following quantities: (a) KEi , (b) KEf , (c)ΔKE, (d) W, and (e) d.

### Solution

(a)KEi = ½ mvi2

= ½ (10. kg)(2.0 m/s)2

= 20. J

(b)KEf = ½ mvf2

= ½ (10. kg)(6.0 m/s)2

= 180 J

### Solution (cont’d)

(c) ΔKE = KEf – KEi

= 180 J – 20. J

= 160 J

(e) W = F∙d

d = W/F

= 160 J / 20. N

= 8.0 m

(d) W = ΔKE

= 160 J

### Gravitational Potential Energy

• An object decreases its gravitational Potential Energy (PE) as it moves closer to the Earth

• To calculate the change in PE of an object we measure the work done on the object. The force needed to overcome gravity is Fg = mg. Therefore, since W = Fg ∙d , PE is defined as

ΔPE = mg Δ h

where Δh represents change in vertical displacement above the earth.

### Example

• A 2.00 kg mass is lifted to a height of 10.0 m above the surface of the Earth. Calculate the change in the PE of the object.

• Solution

ΔPE = mg Δ h

= (2.00 kg) (9.8 m/s2)(10.0m)

= 196 J

### NOTE

• For a change in gravitational energy to occur, there must be a change in the vertical displacement of an object; if it is moved only horizontally, the ΔPE = 0.

• If an object is moved up an inclined plane, its potential energy change is measured by calculating only its verticaldisplacement; the horizontal part does not change its PE

### Home work

• Review book : read pg 84 Do # 26-38 (PE)

• Read pg 92-93 Do #57 -66 (KE)

## Conservation of Mechanical Energy

### Conservation of Mechanical Energy

• In a system, the sum of PE and KE (the total mechanical energy) is constant (i.e. conserved); a change in one is accompanied by an opposite change in the other.

ΔPE = -ΔKE

PEi + KEi = PEf +KEf

### example

• A 0.50 kg ball is projected vertically and rises to a height of 2.0 meters above the ground. Calculate: (a) the increase in the ball’s PE

(b) the decrease in the ball’s KE

(c) the initial KE

(d) the initial speed of the ball

### solution

• The increase in the ball’s PE

ΔPE = mg Δ h

= (0.50 kg)(9.8 m/s2)(2.0 m)

= 9.8 J

(b) The decrease in the ball’s KE

ΔKE = -ΔPE

= -9.8 J

### Solution cont’d

(c) The initial KE

• Recall that at its highest point, the speed of the ball is zero; therefore its KE is zero. So the initial KE represents the change in KE of the object .

ΔKE = KEf – KEi

-9.8J = 0 - KEi

KEi = 9.8 J

### Sol’n cont’d

• The initial speed of the ball

KEi = ½ mvi2

Solving for vi = sqrt (2KEi /m)

= sqrt [ 2(9.8J)/(0.50kg) ]

= 6.3 m/s

### Pendulum

Observe that the falling motion of the bob is accompanied by an increase in speed. As the bob loses height and PE, it gains speed and KE; yet the total of the two forms of mechanical energy is conserved

### example

• A pendulum whose bob weighs 12 N is lifted a vertical height of 0.40 m from its equilibrium position. Calculate:

(a) change in PE between max height and equilibrium height

(b) Gain in KE and

(c) the velocity at the equilibrium point.

### solution

• Take PE at lowest point to be zero

ΔPE = mg Δ h = FgΔ h

= (12 N)(-0.40m)

= -4.8 J

(b) ΔKE = -ΔPE

= -(-4.8 J)

= 4.8 J

### Sol’n cont’d

( c ) First must calculate mass of bob

Fg = mg

m = Fg / g

= 12N / 9.8m/s2

= 1.2 kg

Then calculate velocity

KE= ½ mv2

v = sqrt (2KE/m)

= sqrt [ 2(4.8J)/(1.2kg) ]

= 2.8 m/s

## Elastic Potential Energy and Springs

### Hooke’s Law

• The English scientist Robert Hooke was able to show that the magnitude of a force (F) is directly proportional to the elongation (stretch) of compression of a spring (x) within certain limits.

Fs = kx

k – spring constant – unit is the newton per meter (N/m)

Note: the greater the constant, the stiffer the spring

### Elastic Potential Energy

Spring is attached to a wall. If a force is applied and stretch the string to the right, work has been done. This work has been converted into the spring’s potential energy

– Elastic Potential Energy

PEs = ½ kx2

### Hooke’s law and Potential Energy

Area under the graph equals the work done.

Fs = kx

W

### Example

• A spring whose constant is 2.0 N/m is stretched 0.40 m from its equilibrium position. What is the increase in the elastic potential energy of the spring?

Solution

PEs = ½ kx2

= ½ (2.0 N/m ) (0.40 m)2

= 0.16 J

### Elastic and Inelastic Collisions

• In an elastic collision BOTH kinetic energy and momentum are conserved

p1i + p2i = p1f + p2f

KE1i+ KE2i = KE1f +KE2f

p1i

p2i

p1f

p2f

1

2

1

2

1

2

KE1i

KE2i

KE2f

KE1f

### Inelastic collisions

• In inelastic collisions, the kinetic energy that is “lost” is converted into internal energy (Q) of the objects by frictional forces.

• These systems are called nonideal mechanical systems.

• The energy is constant

ET = PE + KE + Q

A change in the internal energy of an object is usually accompanied by a change in temperature

## Simple Machines and Work

### Simple Machines

• A simple machine is a device that allows work to be done and offers and advantage to the user. Ex: pulleys, levers, inclined planes, wheels and axles and screwdrivers

Win = Wout

Fin ∙ din = Fout ∙ dout

Fout / Fin = din / dout

Mechanical Advantage (MA): Fout / Fin

Ideal MA (IMA) assumes no friction (use din / dout )

Actual MA (AMA) is always less than IMA (use Fout / Fin )

The efficiency of machine is AMA/IMA ratio and this value is always less than 100%

### example

• A 100 N object is moved 2 m up an inclined plane whose end is lifted 0.5 m from the floor. If a force of 50 N is needed to accomplish this task, calculate the (a) IMA

(b) AMA

And (c) efficiency of the inclined plane

### solution

Input force (Fin ) = 50 N (force needed to move object)

Output force (Fout ) = 100N (force that has been lifted)

Input distance (din ) = 2m (distance moved along plane)

Output distance (dout ) = 0.5m (distance object is raised)

• IMA = din / dout = 2m / 0.5m = 4

• AMA = Fout / Fin = 100N / 50 N = 2

• efficiency = AMA/IMA = 2/4 = 0.5 (50%)

## Internal Energy and Work

### Internal Energy (Q)

• Internal Energy of a system is the total kinetic and potential energies of the atoms and molecules that make up the system.

• Recall: a change in the internal energy of an object is usually accompanied by a change in its temperature.

### example

• Force is used to move an object along a horizontal table at constant speed.

• Has work been done?

• Is there a change in Kinetic Energy? Why or why not?

• Is there a change in Potential Energy? Why or why not?

• How was work used?

Yes W = F ∙ d

No – speed is constant

No – table is horizontal - so no change in height

Used to overcome friction between object and table so internal energy of the object-table system has been increased by work done.

## The Laws of Thermodynamics

The study of the relationships among heat, work, and energy in the universe

### The first law of Thermodynamics

• Energy can neither be created nor destroyed. It can only change forms.

• The change in the internal energy of a system (ΔU) is equal to the heat (Q) that the system absorbs (or releases) minus the work (W) it does (or has done on it)

ΔU = Q – W

### 2nd Law of Thermodynamics

• Result if the work of the French physicist Nicolas Carnot with heat engines.

• Law states heat cannot flow from colder object to a warmer one without work being done on the system.

• Ex. Refrigerators must be run by motors in order to withdraw heat from objects placed in them

• No heat engine can be 100% efficient. Some of the heat absorbed by the engine must be lost in the random motion of its molecules.

• Entropy is the measure of this disorder

### 3rd law of Thermodynamics

• As temperature approaches absolute zero, (0 K) the entropy of a system approaches a constant minimum

• The efficiency of a heat engine depends on its operating temperatures; engine would reach 100 % efficiency only at 0 K.

• Engine cannot be completely efficient, therefore 0K cannot be reached.