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# Work and energy

## by guestb32189 on May 09, 2010

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## Work and energyPresentation Transcript

• Topic 2 Work and Energy
• Contents
• Work and Energy
• Kinetic Energy
• Potential Energy
• Elastic Potential Energy
• Conservation of Energy
• Power
• Centripetal Acceleration
• Work and Energy
• Kinetic energy is defined as:
• E k = ½ mv 2
• If a particle is moving freely with no unbalanced force acting on it :
• NI tells us that it will move with constant velocity.
• This means that kinetic energy will also be constant.
• Work and Energy
• What happens however if an unbalanced force acts?
• A constant unbalanced force produces :
• a constant acceleration.
• One of the kinematic equations that can be used in this circumstance is:
• Work and Energy
• v 2 – u 2 = 2 as
• T o find the K.E. multiply both sides by ½m.
• ½ mv 2 - ½ mu 2 = mas
• From NII, F = ma
• ½ mv 2 - ½ mu 2 = Fs
• Work and Energy
•  K.E. = Fs
• The term on the RHS of the equation is called WORK .
• Work and Energy
• The work done by a constant unbalanced force acting on a particle :
• which is moving in one dimension is given by ,
• the product of the unbalanced force and ,
• the displacement produced.
• W = Fs
• Work and Energy
• This equation shows us that if an unbalanced force acts :
• there will always be a change in kinetic energy and ,
• an amount of work done.
• A glider moving at constant velocity on an air track has :
• no unbalanced force acting on it.
• Work and Energy
• However, if it is on a slope ;
• t here is an unbalanced force ,
• of gravity (weight),
• acting on it and it will accelerate.
• This weight can be resolved into two components,
• parallel and perpendicular to the motion.
• Work and Energy
• Work and Energy
• The perpendicular component of the weight :
• is balanced by the reaction force ,
• of the air track on the glider ,
• air on the glider.
• Work and Energy
• The unbalanced force is therefore the parallel component of the weight.
• This force :
• multiplied by the displacement along the track gives ,
• the work done on the glider.
• Work and Energy
• What part does the angle of inclination play in calculating the work done?
• Example
• A Woolworths supermarket trolley (that does move in the direction you push it), is pushed with a force of 200 N acting at an angle of 40 o to the ground. Find the effective horizontal force pushing the trolley along.
• Solution
•  = 40 o
• I F I = 200 N
• Draw vector diagram
• Solution
• Solution
• F H is the effective force pushing the trolley
• F H = F cos 
• F H = 200 x cos 40 o
• F H = 200 x 0.7660444
• F H = 153 N Horizontally
• Work and Energy
• Work can be determined by studying a force-displacement graph.
Force (N) Displacement (m) 10 5
• Work and Energy
• Area under graph = height x length
• Area under graph = Force x displacement
• Force x displacement = Work
• Area under graph = Work done
• Area under graph = 5 x 10
• Area under graph = 50 J
• Work and Energy
• Work is easy to calculate when the force is constant.
• What happens if the force is not constant?
• Use a F vs. disp. graph.
• Work and Energy Force (N) Displacement (m) 10 5
• Work and Energy
• Work = Area under a F vs. Disp. Graph
• Work = ½ (b x h)
• Work = ½ (5 x 10)
• Work = 25 J
• Energy and Power
• Kinetic Energy
• Push an object and it can move.
• If an object moves:
• it is capable of doing work.
• The object has energy associated with its motion called:
• Kinetic Energy
• Energy and Power
• W = Fs
• F = ma
• W = mas
• v 2 – u 2 = 2as
• Energy and Power
• As W = mas
• W = ½mv 2 – ½mu 2
• W =  ½mv 2
• The quantity  ½mv 2 is called:
• Kinetic Energy
• Energy and Power
• Kinetic Energy, E k , can be defined as:
• The product of half the object’s mass m ,
• and the square of its speed v .
• Energy and Power
• Potential Energy
• Kinetic energy is the ‘energy of motion’.
• We can develop an expression for the energy that is dependent on position;
• potential energy.
• Energy and Power
• Consider an object that is dropped from a height above the floor, h t :
• where the floor is at height h o .
• Displacement is given by s = h t - h o .
• The unbalanced force is given by :
• the weight of the object m g .
• Energy and Power
• As W = Fs
• W = m g ( h t - h o )
• or W = m g  h
• This gives the work done in terms of the objects position.
• This quantity mgh , is defined as the gravitational potential energy.
• Energy and Power
• P.E. = m gh
• Work can also be defined as :
• the change in gravitational potential energy.
• When an object falls :
• it loses gravitational potential energy,
• and gains kinetic energy.
• Energy and Power
• Work can be calculated by the change in either of these two terms.
• Generally, work is defined as the change in energy.
• Energy and Power
• The relationship between E k , E p and work can be shown using a downhill skier.
• Energy and Power
• Energy transformation can be shown using a roller coaster.
• Energy and Power
• Elastic Potential Energy
• Consider a spring that has been compressed.
• When released for time t ,
• the uncompressed position.
• Energy and Power
• This means there must be an unbalanced force acting.
• This force is given by Hooke’s Law.
• The restoring force in a spring is :
• proportional to its extension or compression.
• Graphically, it can be described as:
• Energy and Power
• Energy and Power
• Mathematically, it can be described as:
• F = - kx
• Where k is the slope of the graph.
• Energy and Power
• The elastic potential energy can also be calculated.
• E.P.E. = ½ kx 2 .
• This suggests that as a spring is compressed or extended :
• the energy increases.
• Energy and Power
• Conservation of Energy
• Consider a ball thrown vertically into the air.
• It begins its motion with kinetic energy.
• As it reaches it’s highest point :
• The E k is zero.
• Energy and Power
• At the same time, the G.P.E. has :
• increased.
• The loss of one type of energy :
• is balanced by the gain in another.
• Total Energy = mgh + ½mv 2 .
• If a glass of whisky is pushed along a bar to a waiting gunslinger :
• is energy conserved ?
• Energy and Power
• In this case,
• the G.P.E. has not increased :
• when the K.E. has decreased.
• This however is not an isolated system.
• Energy has been lost to friction.
• The total energy in any isolated system :
• is constant .
• Energy and Power
• A dart is fired out of a gun using a spring.
• Energy and Power
• A 3 kg cart moves down the hill.
• Calculate the E p lost and E k gained.
• Energy and Power
• E p = mgh
• E p = 3 x 9.8 x (0.40 – 0.05)
• E p = 10.3 J
• E k = ½ mv 2
• E k = ½ x 3 x 2.62 2
• E k = 10.3 J
• Energy is conserved.
• Energy and Power
• Energy can be expended to perform a useful function.
• A device that turns energy into some useful form of work is called a:
• Machine
• Energy and Power
• Machines cannot turn all the energy used to run the machine into useful work.
• In any machine, some energy goes to:
• atomic or molecular kinetic energy.
• This makes the machine warmer.
• Energy is dissipated as heat.
• Energy and Power
• The amount of energy converted into:
• useful work by the machine is called,
• The efficiency .
• An example of a simple machine is:
• A pulley system.
• We can do 100 J of work.
• Energy and Power
• Friction turn the pulleys which in turn rub on the axles.
• This may dissipate 40 J of energy as heat.
• The system is 60% efficient.
• Energy and Power
• Efficiency can be expressed mathematically:
• Energy and Power
• Power
• Power is defined as :
• the rate at which work is done .
• Units:
• Js -1 or Watts.
• Energy and Power
• The work in this equation could be :
• the change in kinetic energy or ,
• the work done on a mass that has been lifted.
• It does not matter what form the energy takes :
• it is just the rate at which work is done.
• Energy and Power
• A 100W light globe produces 100 J of energy every second.
• To give an idea of the size of 1 W,
• a jumping flea produces 10 -4 W,
• a person walking 300 W and ,
• a small car 40 000 W.
• Uniform Circular Motion
• Centripetal Acceleration
• A particle undergoing uniform circular motion is continually changing velocity :
•  acceleration is changing.
• Uniform Circular Motion
• Uniform Circular Motion
•  v 1 = v b - v a
•  v 2 = v c - v b and so on
• The magnitude of  v 1 =  v 2
• The direction is always to the centre of the circle .
• Uniform Circular Motion
• The acceleration,
• which produces these velocity changes in a direction which is ,
• always towards the centre of the circular motion, is called :
• centripetal (centre seeking) acceleration.
• Uniform Circular Motion
• Newton’s 2 nd law tells us that :
• a centripetal acceleration can only happen if ,
• there is an unbalanced force.
• Uniform Circular Motion
• Any particle undergoing uniform circular motion is acted upon by :
• an unbalanced force which is ,
• constant in magnitude and ,
• directed towards the centre of the circle.
• This is called Centripetal Force .
• Uniform Circular Motion
• Uniform Circular Motion
• With a centripetal force, the object moves in a circular path.
• Uniform Circular Motion
• When the unbalanced force is released :
• the object moves along a tangential path ,
• at a constant velocity .
• Uniform Circular Motion
• Examples include:
• Moon revolving around the Earth:
• Gravitational Force,
• Directed towards the centre of the Earth,
• Holds the moon in a near circular orbit.
• Uniform Circular Motion
• Electrons revolve around the nucleus:
• Electric Force,
• Directed to centre of the nucleus,
• Holds electrons in circular orbit
• Uniform Circular Motion
• Car rounding a corner:
• Sideways frictional force,
• Directed towards centre of turn,
• Force between car tyre and road.
• If force not great enough:
• Car skids.
• Uniform Circular Motion
• The force acts on the passenger in the car if they do not have their seat belt on.
• Note: it is an European car
• Uniform Circular Motion
• Washing Machine tub on spin cycle:
• Tub rotates at high speed,
• Inner wall exerts inwards force on clothes.
• Holes in tub allow water to follow a straight line.
• Water escapes.
• Force acts on clothes:
• not water.