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# Work, energy & power physics

## on Mar 10, 2014

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## Work, energy & power physics Presentation Transcript

• Work, EnErgy & PoWEr Objectives: to begin developing a concept of energy – what it is, how it is transformed and transferred. show an understanding of the concept of work in terms of the product of a force and displacement in the direction of the force
• Work, EnErgy & PoWEr • Objectives:  calculate the work done in a number of situations including the work done by a gas that is expanding against a constant external pressure: W = p ΔV
• Concept of Energy Work, energy and power
• Concept of Energy Work, energy and power
• Concept of Energy An analogy: ability to purchase good and services = ENERGY ability to do work (simple but inadequate) Work, energy and power
• Concept of Energy Kinetic energy is analogous to cash just like energy, can be transformed and transferred in various ways Potential energy is analogous to savings account Work is analogous to getting a paycheck (+) or paying bills (making purchases) ( - ) Work, energy and power
• Work, EnErgy & PoWEr So what then is WORK? -it is the mechanical transfer of energy to or from the system by the pushes and pull of external forces - it is the product of force and distance in the same direction of the force.
• Work, EnErgy & PoWEr Calculating work done 2 factors are involved External force, F must be applied S = distance moved Object must move in the direction of the force, s
• Work, EnErgy & PoWEr W=Fxs The work done by a force is defined as the product of the force and the distance moved in the direction of the force.  the magnitude of the force F – the bigger the force, the greater the amount of work you do.  the distance s you push the car – the further you push it, the greater the amount of work done.
• Work, EnErgy & PoWEr work done W = F x s = 300 x 5.0 = 1500 J = 1.5 kJ S = distance moved F = 300 N s = 5.0 m
• Work, EnErgy & PoWEr Doing work Pushing a car to start it moving: your force transfers energy to the car. The car’s K.E. (i.e. movement energy) increases. Lifting weights: you are doing work as the weights move upwards. The GPE of the weights increases. Not doing work Pushing a car but it does not budge: no energy is transferred, because your force does not move. The car’s K.E. does not change. Holding the weights above your head: you are not doing work on the weights because the force you apply on them is not causing it to move. The GPE of the weights is not changing Writing an essay: you are doing work Reading an essay: this may seem like because you need a force to move your “hard work”, but no force is involved, pen across the page, or to press the so you are not doing any work. keys on the keyboard
• Work, EnErgy & PoWEr work done = energy transferred Doing work is a way of transferring energy. 1 Joule = 1 newton metre 1J=1Nm The Joule is defined as the amount of work done when a force of 1 Newton moves an object a distance of 1 metre in the direction of the force
• Work, EnErgy & PoWEr work done = energy transferred The Joule is also defined as the amount of energy transferred when a force of 1 Newton moves an object a distance of 1 metre in the direction of the force
• Work, EnErgy & PoWEr Test yourself 1. In each of the following examples, explain whether or not any work is done by the force mentioned. a) You pull a heavy sack along the ground. YES b) The force of gravity pulls you downwards when you fall.YES c) The tension in a string pulls on a stone when you whirl it around at a steady speed. NO d) The contact force of the bedroom floor stops you from falling into the room below. NO 2. A man of mass 70 kg climbs stairs of vertical height 2.5 m. Calculate the work done against the force of gravity. 1716.75 = 1.7 kJ
• Work, EnErgy & PoWEr Test yourself 3. A stone of weight 10 N falls from the top of a 250 m high cliff. a) Calculate how much work is done by the force of gravity in pulling the stone to the foot of the hill. 2500 J b) How much energy is transferred to the stone? 2500 J
• Work, EnErgy & PoWEr Force, distance and direction For force to do work, there must be movement in the direction of the force
• Work, EnErgy & PoWEr Fx = F cos 30o = 50 cos 30o = 43. 301270189 horizontal component (Fx) of 50 N Work done = (Fx) x s = 43. 301270189 x 10 = 433.01270189 = 4.3 x 102 J
• Work, EnErgy & PoWEr Doing work against gravity A 80.0 kg man is climbing a stair as shown in the diagram on the right. Given the dimensions calculate for the work done by the man upon reaching the last step above.
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• Work, EnErgy & PoWEr Quick review: Doing/Not Doing work???? F d
• Work, EnErgy & PoWEr F = 0 12 N Θ = 35o Fx = ? An airport passenger is pulling his luggage with a force F = 120 N along a level floor. The pulling force makes an angle of 35o measured from the floor and the passenger moves through a distance d of 6.0 m, what is the work done on the luggage? d = 6.0 m
• Work, EnErgy & PoWEr A stunt person slides down a cable that is attached between a tall building and the ground. The stunt person has a mass of 85 kg. The speed of the person when reaching the ground is 20 m s−1. Calculate: a)the change in gravitational potential energy of the person b) the final kinetic energy of the person c) the work done against friction d) the average friction acting on the person.
• Work, EnErgy & PoWEr The figure on the right shows the forces acting on a box which is being pushed up a slope. 100 N 70 N 30 N 100 N 45o Calculate the work done by each force if the box moves up 0.5 m up the slope.
• Work, EnErgy & PoWEr Try these: 1. Calculate how much gravitational potential energy (GPE) is gained if you climb a flight of stairs. Assume that you have a mass of 52 kg and that the height you lift yourself is 2.5 m. 2. A climber of mass 100 kg (including all the equipment he is carrying) ascends from sea level to the top of a mountain 5500 m high. Calculate the change in her gravitational potential energy (GPE)
• Work, EnErgy & PoWEr 3. Calculate the increase in kinetic energy of a mass 800 kg when it accelerates from 20 m/s to 30 m/s. Change in KE = 200 kJ 4. Calculate the change in KE of a ball of mass 200 g when it bounces. Assume that it hits the ground with a speed of 15.8 m/s and leaves it at 12.2 m/s. Change in KE = 10 J
• Work, EnErgy & PoWEr movable piston Work done by gas P= force _____ area force exerted by the gas when it expands F=p A work done is W=pAs A s ??? Change in volume, ΔV W=pΔV
• Work, EnErgy & PoWEr When the gas EXPANDS, work is done BY the gas. When the gas CONTRACTS, then work is done ON the gas.
• Work, EnErgy & PoWEr Study the sample problems on pages 76, 80, & 82 and answer the “Now it’s your turn” questions. Check your answers against the answers from the book.
• Work, EnErgy & PoWEr The diagram shows a child on a swing. The mass of the child is 35 kg. The child is raised to point A and then released. She swings downwards through point B. a) Calculate the change in gravitational potential energy of the child between A and B. b) Assuming that air resistance is negligible, calculate the speed of the child as she passes through the equilibrium position B. c) The rope stays taut throughout. Explain why the work done by the tension in the rope is zero.
• Work, EnErgy & PoWEr A bullet of mass 30 g and travelling at a speed of 200 m s−1 embeds itself in a wooden block. The bullet penetrates a distance of 12 cm into the wood. Using the concepts of work done by a force and kinetic energy, determine the average resistive force acting on the bullet. Work done by resistive force = initial kinetic energy of bullet 1 × 0.030 × 200 2 2 0.030 × 200 2 F= = 5.0 × 10 3 N (5.0 kN) 2 × 0.12 F × 0.12 =
• Work, EnErgy & PoWEr The diagram shows a 50 kg crate being dragged by a cable up a ramp that makes an angle of 24° with the horizontal. The crate moves up the ramp at a constant speed and travels a total distance of 20 m up the ramp. Determine the magnitude of the friction between the crate and the surface of the ramp.
• Work, EnErgy & PoWEr Gain in gravitational potential energy = mgh = 50 × 9.81 × 20 sin 24° = 3.99 × 103 J Work done by tension in the rope = F cos θ × x = 350 × cos 30° × 20 = 6.06 × 103 J
• Work, EnErgy & PoWEr Work done against friction = 6.06 × 103 − 3.99 × 103 = 2.07 × 103 J Since work done is given by: work done = force × distance we have: 3 friction 2.07 × 10 = = 104 N ≈ 100 N 20
• Work, EnErgy & PoWEr An object of mass m passes a point X with a velocity v and slides up a frictionless incline to stop at point Y which is at a height h above X v Y h X A second ball of mass 0.5m passes X with a velocity of 0.5v. To what height will it rise?
• Work, EnErgy & PoWEr 10.0 m 1.0 m A body of mass 1.0 kg initially at rest slides down an incline plane that is 1.0 m high and 10.0 m long. If the body experiences a constant resistive force of 0.5 N over the slope, what is the KE of the body at the base of the plane? Gain in KE = loss in PE – Work against resistive force 4.8J
• Work, EnErgy & PoWEr Power work done Power = time taken Watts (W) - is the rate of working 1 joule per second ( scalar just like energy ) J s
• Work, EnErgy & PoWEr It’s common to say that a strong person is “powerful” In Physics, strength, or force, and power are NOT the same. Large forces may be exerted w/o any movement and thus NO WORK is done and the power is zero. Large rock resting on the ground is not moving and yet exerts a large amount of force.
• Work, EnErgy & PoWEr Consider a force F which moves a distance x at constant speed v in the direction of the force, in time t W=Fs W=Fs t t Dividing both sides by t but W/t = power and s/t = speed P=Fv Power = force x speed
• Work, EnErgy & PoWEr Study the sample problem from page 103, then solve “Now it’s your turn” problems 1-3
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