Work, energy & power physics

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


Concept of Energy


An analogy:

= ENERGY

ability to do work (simple

but inadequate)

ability to purchase good and services

Concept of Energy


Kinetic energy is analogous to cash

Potential energy is analogous to savings account

Work is analogous to getting a paycheck (+) or paying bills (making purchases) ( - )

just like energy, can be transformed and transferred in various ways


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.


Calculating work done2 factors are involved

S = distance moved

External force, F must be applied

Object must move in the direction of the force, s


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.

W = F x s The work done by a force is defined as the product of the force and the distance moved in the direction of the force.


S = distance movedF = 300 Ns = 5.0 m

work done W = F x s = 300 x 5.0 = 1500 J = 1.5 kJ


Doing work Not 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.

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.

Lifting weights: you are doing work as the weights move upwards. The GPE of the weights increases.

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 because you need a force to move your pen across the page, or to press the keys on the keyboard

Reading an essay: this may seem like “hard work”, but no force is involved, so you are not doing any work.


work done = energy transferred

Doing work is a way of transferring energy.

1 Joule = 1 newton metre1 J = 1 N m

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 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 & PowerTest 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.b) The force of gravity pulls you downwards when you fall.c) The tension in a string pulls on a stone when you whirl it

around at a steady speed.d) The contact force of the bedroom floor stops you from

falling into the room below.

2. A man of mass 70 kg climbs stairs of vertical height 2.5 m. Calculate the work done against the force of gravity.

YESYES

NO

NO

1716.75 = 1.7 kJ

Work, Energy & PowerTest 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.

b) How much energy is transferred to the stone?

2500 J

2500 J


Force, distance and direction

For force to do work, there must be movement in the direction of the force


horizontal component (Fx) of 50 N

Fx = F cos 30o = 50 cos 30o

= 43. 301270189

Work done = (Fx) x s = 43. 301270189 x 10 = 433.01270189 = 4.3 x 102 J


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.


Work, Energy & PowerQuick review: Doing/Not Doing work????

F

d


Θ = 35o

F = 120 N

Fx = ?

d = 6.0 m

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?

Work, Energy & PowerA 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 personc) the work done against frictiond) the average friction acting on the person.


45o

100 N70 N

100 N

30 N

The figure on the right shows the forces acting on a box which is being pushed up a slope.

Calculate the work done by each force if the box moves up 0.5 m up the slope.


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)


3. Calculate the increase in kinetic energy of a mass 800 kg when it accelerates from 20 m/s to 30 m/s.

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

Change in KE = 200 kJ


force exerted by the gas when it expands

F = p Awork done is

W = p A s

movable piston

P = forcearea

_____

A s ???Change in

volume, ΔV

W = p Δ V

Work done by gas


When the gas EXPANDS, work is done BY the gas.

When the gas CONTRACTS, then work is done ON the gas.


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 & PowerThe 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.


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

2200030.02

112.0 F

kN) (5.0 N 100.512.02

200030.0 32

F

Work, Energy & PowerThe 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 done by tension in the rope F cos x

350 cos 30° 20

6.06 103 J

Gain in gravitational potential energy mgh

50 9.81 20 sin 24°

3.99 103 J


Work done against friction 6.06 103 3.99 103 2.07 103 J

Since work done is given by:work done force distancewe have:friction N 100N 104

20

1007.2 3


h

v Y

X

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

A second ball of mass 0.5m passes X with a velocity of 0.5v. To what height will it rise?


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 force4.8J


Power

Power = work done time taken

Js

Watts (W)

- is the rate of working 1 joule per second ( scalar just like energy )


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.


Consider a force F which moves a distance x at constant speed v in the direction of the force, in time t

W = F sDividing both sides by t

W = F st t

but W/t = power

and s/t = speed

P = F v

Power = force x speed


Study the sample problem from page 103, then solve “Now it’s your turn” problems 1-3













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Work, energy & power physics