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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)

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definitions
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
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
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

example1
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?
definition
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).

example2
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
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

slide10
Hw
  • Read pg. 80-83 Do # 1-25
energy

Energy

Mechanical Energy

kinetic 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

example3
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
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
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
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.

example4
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

slide18
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
Home work
  • Review book : read pg 84 Do # 26-38 (PE)
    • Read pg 92-93 Do #57 -66 (KE)
conservation of mechanical energy1
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

example5
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

solution1
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 d1
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
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
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

example6
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.

solution2
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 d1
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

hooke s law
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
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
Hooke’s law and Potential Energy

Area under the graph equals the work done.

Fs = kx

W

example7
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
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
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
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%

example8
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

solution3
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 q
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.
example9
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 Laws of Thermodynamics

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

the first law of thermodynamics
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

2 nd law of thermodynamics
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
3 rd law of thermodynamics
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.
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