Power, Efficiency, & Energy

November 2012

Power

Power is the work done in unit time or energy converted in unit time

measures how fast work is done or how quickly energy is converted.

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orW E

P = P = t t

Power

Power is the work done in unit time or energy converted in unit time

measures how fast work is done or how quickly energy is converted.

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Units: ◘1J(joule)

1 W(Watt) = 1s

A 100 W light bulb converts electrical energy to heat and light at the rate of 100 J every second.

orW E

P = P = t t

Calculate the power of a worker in a supermarket who stacks shelves 1.5 m high with cartons of orange juice, each of mass 6.0 kg, at the rate of 30 cartons per minute.

Calculate the power of a worker in a supermarket who stacks shelves 1.5 m high with cartons of orange juice, each of mass 6.0 kg, at the rate of 30 cartons per minute.

0W Fdcos0 (30×60N)×1.5mP = = =

t t 60s P = 45 W

Efficiency

Efficiency is the ratio of how much work, energy or power we get out of a system compared to how much is put in.

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useful outputefficiency =

total input

No units◘

Efficiency can be expressed as percentage by multiplying by 100%.

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out out out

in in in

W E Peff = = =

W E P

No real machine can ever be 100% efficient, because there will always be some energy lost as heat

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A car engine has an efficiency of 20 % and produces an average of 25 kJ of useful work per second.How much energy is converted into heat per second.

A car engine has an efficiency of 20 % and produces an average of 25 kJ of useful work per second.How much energy is converted into heat per second.

out

in

Eeff =

E

in

25000J0.2 =

EEin = 125000 J

heat = 125 kJ – 25 kJ = 100 kJ

Quick Review (2 min)

With your table partner …

Person with longer hair:• Define power in your own words• State the equation and units for power.

Person with shorter hair:• Define efficiency in your own words• State the equation and units for efficiency

Energy

Energy = the potential to do work Units: Joules

And, in reverse… work done = change in energy W = ∆ E

Energy

Energy = the potential to do work Units: Joules

And, in reverse… work done = change in energy

Two major types of energy1) Potential energy – energy that is stored due to an

object’s position.Examples: gravitational potential energy

elastic potential energy2) Kinetic energy - energy of motion

W = ∆ E

Gravitational Potential Energy

Stored energy due to an object’s height.The higher the object, the more gravitational potential energy.**

PEg = Fgd = mgh PEg = mgh

Elastic Potential Energy

Energy stored due the compression or stretching of elastic material, such as a spring.

Fspring = kx (Hooke’s Law)

where k is the spring constant and x is the distance the spring has been stretched or compressed.

Pespring = ½ kx2

Potential Energy and Work

We said earlier that W = Δ E

and we know that potential energy depends only on the position of an object…

THIS IMPLIES THAT WHEN DOING WORK, THE PATH DOESN’T MATTER --- ONLY THE END POSITION.

Work doesn’t depend on path.Imagine a mass m lifted a vertical distance h following two different paths. 1. straight up:

–– the work done against gravitational force is:

W = Fg h cos 00

W = mg h

2. along a ramp a distance d (no friction) – minimum force needed is F2 = mg sinθ. – the work done against gravitational

force is:

W = F2 d cos 00 = mg sinθ d

W = mg h

hd

mg

F2

θ

W depends on VERTICAL distance moved, not on the path

Work doesn’t depend on path.Imagine a mass m lifted a vertical distance h following two different paths. 1. straight up:

–– the work done against gravitational force is:

W = Fg h cos 00

W = mg h

2. along a ramp a distance d (no friction) – minimum force needed is F2 = mg sinθ. – the work done against gravitational

force is:

W = F2 d cos 00 = mg sinθ d

W = mg h

hd

mg

F2

θ

Also notice: W = the potential energy of the block!

A ramp reduces the force needed – makes life easier

1. W = F h = mg h

2. W = F d = mg sinθ h/sinθ = mg h

θ

hhd

W = F h or F d (along the ramp)

Work and PE – You do

a) A cart filled with bricks is pulled at constant speed up a ramp. The cart has a mass of 3.0kg and the heightof the ramp is 0.45 m.What is the PEg of the cart at

the top of the ramp?

b) It required 14.7 N to pull the cart up the ramp, and the ramp is 0.9 m long. How much work was needed to pull the cart up the ramp?

Work and PE – You do

a) A cart filled with bricks is pulled at constant speed up a ramp. The cart has a mass of 3.0kg and the heightof the ramp is 0.45 m.What is the PEg of the cart at

the top of the ramp?

b) It required 14.7 N to pull the cart up the ramp, and the ramp is 0.9 m long. How much work was needed to pull the cart up the ramp?

Both answers are 13.2 J!

Kinetic energy

Kinetic energy is the energy of motion.

KE = ½ m v2 where v = velocity

This means that a moving object possesses the capacity to do work.

Example: A hammer by virtue of its motion can be used to do work in driving a nail into a piece of wood.

(1) A father pushes his child on a sled on level ice, a distance 5 m from rest, giving a final speed of 2 m/s. If the mass of the child and sled is 30 kg, how much work did he do?

W = ∆ KE = ½ m v2 – 0 = ½ (30 kg)(2)2 = 60 J

(2) What is the average force he exerted on the child?

W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N

(3)A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force.

How much work must this friction force do in order to stop the car?

d

vi = 45 km/h

Ffr

(work done by friction force)

vf = 0

(3)A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force.

How much work must this friction force do in order to stop the car?

W = ∆ KE = 0 – ½ m u2 = – ½ (1000 kg) (45 x1000 m/3600 s)2

W = – 78125 J = – 78 kJ

(the – sign just means the work leads to a decrease in KE)

d

vi = 45 km/hr

Ffr

(work done by friction force)

vf = 0

(4) Instead of slamming on the brakes the work required to stop the car is provided by a tree!!! What average force is required to stop a 1000-kg car going at 45 km/h if the

car collapses one foot (0.3 m) upon impact?

corresponds to 29 tons hitting you OUCH

Do you see why the cars should not be rigid. Smaller collapse distance, gretaer force, greater acceleration. For half the distance force would double!!!! OUCH, OUCH

The net force acting on the car is F

W = – 78125 J

W = – F d = – ½ m u2

F = 260 x 103 N

a = (v2 - u2 )/2d = - 520 m/s2 HUGE!!!!

- W 78125F = =

d 0.3

KE and Work – You do Determine the KE of a 625 kg roller coaster

that is moving with a speed of 18.3 m/s.

What would the KE be if the roller coaster doubled its speed?

KE and Work – You do Determine the KE of a 625 kg roller coaster

that is moving with a speed of 18.3 m/s. 1.05 X 105

J

What would the KE be if the roller coaster doubled its speed?

4.19 X 105 J