Work, Power, & Efficiency

Work• Work is done when a force is exerted

on an object and the object moves a distance x.Work = force ·distance

W =F· xUnit: N·m = J (called a Joule)Work is a scalar quantity.

WorkWhen you hold something, are you exerting

a force on the object? Yes.When you hold something, are you doing

work? No.If you set the object on a table, does the

table exert a force on the object? Yes. Does the table do any work? No.

When you hold something, then, you are not doing work either.

Work

Work

• Work is only done by a force on an object if the force causes the object to move in the direction of the force.

Work• When a force is

applied at an angle, use the component of the force in the direction of motion.

xΔθcosFW

Work• For work done by the frictional force:

• The work is negative because friction acts in a direction opposite to the motion x.

xΔFW f

gmμFf

Work• For an object moving at constant speed

(a = 0 m/s2), Fx = Ff.

• If the force and distance are parallel ( = 0°; cos 0 = 1), the amount of work is positive, but if the two vectors are anti-parallel ( = 180°; cos 180 = -1), then the work is negative.

9

When the displacement is in the same direction as the force, the work is positive.

When the displacement is in the opposite direction of the force, the work is negative.

Work Done by a Gravitational Force• One constant force we already have

dealt with is the force of gravity.• So we should be able to compute the

work done on an object as it rises and falls…

• So lets look at a particle-like tomato of mass m that is thrown upward with an initial velocity v0.

• As the tomato rises, work is done on the tomato by the force of gravity.

Work Done by a Gravitational Force

• The work done is:• Because the tomato is rising, the

force is opposite in direction to the displacement ( = 180º; cos 180 = -1), thus the work done is negative.

• As we expect, the tomato’s velocity decreases until it reaches zero; it then “turns around” and begins to fall.

W = m g d cos θ

W = -m g d

Work Done by a Gravitational Force

• At this time, the force is in the same direction as the displacement = 0º; cos 0 = 1), so the work done is positive.

W = m g d cos θ

W = m g d

Work Done in Liftingand Lowering an Object

• When we lift an object, our applied force does positive work Wa on an object while at the same time gravity is doing negative work Wg on the object:

• This also applies when an object is lowered.

appliedW = m g d

gravityW = -m g d

Work Done in Liftingand Lowering an Object

• So if the displacement is vertically upward, the work done is:

• If the displacement is vertically downward, then the work done is:

appliedW = m g d

gravityW = -m g d

Work Done in Liftingand Lowering an Object

• Notice that the work done on the object:

is independent of the magnitude of the force exerted on the object.

W = ±m g d

Work and Inclined Planes• Fx = gravitational component

• W = Fx·x

• The direction of motion is positive.– Fx is negative if the object moves up the

incline.– Fx is positive if the object

moves down the incline.

θsinFF

θsingmF

wx

x

Work and Inclined Planes• For masses on a horizontal surface or

on an incline, the work done by the normal force Fn = 0 J.

• The normal force is not in the direction of the motion, therefore, x = 0 m.

Work and Inclined Planes• Refer to the problem to determine if

you have to worry about friction. If frictionless, set Ff = 0 N.

Elastic Objects/The Spring Force• We know from common experience that the

harder you push on a spring, the harder it pushes back; the same is true for pulling…

• The force needed to stretch or compress an elastic object is a varying force, increasing with increasing distance from the equilibrium position (the unstretched position, where x = 0).

Spring (Elastic) Constant k• F = -k·x (this equation is called Hooke’s law) x = distance of stretch/compression• k = spring constant; unit = N/m.• The spring constant is a measure of the stiffness of

the spring; the larger k is, the stiffer the spring is (and the more force is needed for a given displacement)

• Negative sign indicates that the force F of the spring is in the opposite direction to the distorting force (the pull or the push on the spring). In the problems you will solve, the negative sign is often dropped.

• The force F is always directed toward the equilibrium position.

• Work = 0.5·k· x2

ΔxF

k

Springs in Parallel• Each spring will

share the weight Fw and stretch the same distance x.

• Effective spring constant = k1 + k2

xΔF

kk w21

Springs in Series• Each spring supports the

entire weight Fw and will stretch independently.

2

w2

2

w2

1

w1

1

w1

kF

xΔxΔ

Fk

kF

xΔxΔ

Fk

Springs in Series• Total distance of stretch = x1 + x2

• Effective spring constant:

21

w

xΔxΔF

k

Machines1. Pulleys2. Levers3. Inclined planes4. Wheel and axle5. Wedge6. Screw

• Lever Applet• Pulley Applet

Machines• Machines can change the direction of a

force and/or multiply the force.

• Work comes in two forms: input and output– Work input (Win) = Feffort·xeffort

– Work output (Wout) = Fresistance·xresistance

Machines• Ex: push down on a lever to raise a load Fw;

the weight Fw is the resistance force.

• Work input (Win) is what you do; the effort force Fe that you apply to the machine and the distance this force moves xe.

Machines• Work output (Wout) is what gets done;

the resistance force FR that is moved by the machine and the distance this force moves xR.

• Wout cannot be greater than Win due to friction.

Machines• Ex: Inclined planes require only a

force equal to the parallel component of the weight (Fx), but this force has to be applied over a longer distance L.

hFwhFW

LFLFW

Rout

xein

Mechanical Advantage• Mechanical Advantage (MA): indicates the

amount by which the effort force Fe is increased by the machine.

Dimensionless(no units)

• Ex: if the MA of a machine is 4, the machine multiplies Fe by 4 to get FR; if Fe = 10 N, then FR = 4·10 N = 40 N.

• MA does take friction into account.• The MA of a pulley is equal to the number

of ropes that support weight.

e

R

FF

MA

• There is a trade off between force and distance; if it takes less force, the force has to move through a greater distance.

Ideal Mechanical Advantage

• Ideal Mechanical Advantage (IMA): the mechanical advantage of a machine assuming 100% efficiency – no loss of force to overcoming friction.

Dimensionless(no units)R

e

xx

IMA

Efficiency• Efficiency: the ratio of work output

to work input.

IMAMA

e%100Win

Woute

%100xFhF

e:lifting%100xFxF

eee

w

ee

RR

Efficiency• Efficiency is always less than 1 (as a

decimal) or 100% due to frictional losses.

• The higher the efficiency, more of the work input is converted to work output; less of the work output is lost to overcoming friction.

Power• Power: the rate at which work is

done.

• Unit: J/s = W (watt)• 1 kilowatt (kW) = 1000 W

tW

P

Power• Work = force·distance• Velocity = distance/time

• Horsepower (hp): power produced by a typical British working horse.

• 1 hp = 746 W• horsepower in a car engine:

vFtxF

tW

P

5252rpmtorque

hp

Things To Look For• Pulling an object with friction: flat surface

• Work done by the pulling force F: W = Fx·xWork done by friction: W = -Ff·x

nf

ywn

x

y

FμF

FFF

θcosFF

θsinFF

Things To Look For• Pushing an object with friction: flat surface

• Work done by the pushing force F: W = Fx·xWork done by friction: W = -Ff·x

nf

ywn

x

y

FμF

FFF

θcosFF

θsinFF

Things To Look For• Pushing an object up an incline with

friction: at constant speed (a = 0 m/s2)

xFW

FFF

FμF

θsinFF

θcosFFF

a

fxa

nf

wx

wyn

Things To Look For• If there is an acceleration up the incline

W = (Fx + Ff + m·a)·x

• For an elevator problem with friction:Fup = Fw + Ff = m·g + Ff

P = Fup·v

• If there is an acceleration: Fup = Fw + Ff + m·a = m·g + Ff + m·a

P = Fup·v