Work and energy
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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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Work and Energy

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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)

    (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.


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