WorkWork

- Work (W) is defined as a force moved over a distance

- Only the component of the force in the direction of motion does work

Units: N m

The cart (above) is pulled at constant speed with a force of 20N over a distance of 15m. Determine the work done by the applied force if the handle is pulled a) in a straight line and b) at an angle of 530 ,

a) F = 20 N d = 15 m

W = FH d

W = (20N)(15m)

W = 300 Nm

3

4

5530

370

b) FH / F = 3 / 5

FH = (3 / 5) F = (3 / 5) (20N) = 12 N

W = FH d = (12N)(15m)

W = 180 Nm

Transformation of EnergyTransformation of Energy

When work is done, energy is transformed from one form into another

Consider a planet moving in an elliptical orbit around the sun

v

v

v

v

Fg

Fg Fg

Fg

No work

No energy change

Work done slowing down planet

Energy changes from kinetic to GPE

Work done increasing the planet’s speed

Energy changes from GPE to kinetic

No work

No energy change

EnergyEnergy

- Energy (E) is defined as the capacity to do work Units: Joule (J)

-Energy is the conceptual system for explaining how the universe works and accounting for changes in matter

1 Calorie (C) = 1 kcal = 4186 J

-There are many types of energy which are divided up into mechanical and non-mechanical forms

Form of Non-Form of Non-Mechanical Mechanical

EnergyEnergyAssociated with…Associated with…

Chemical

Thermal

Nuclear

Electromagnetic

bonds between atoms

vibration of atoms

bonds between protons and neutrons in nucleus

Vibration of electric charges

Form of Form of Mechanical Mechanical

EnergyEnergyAssociated with…Associated with…

Kinetic an object that is moving

Gravitational Potential

an object’s position in a gravitational field

Elastic Potential

stretched or compressed elastic materials

Spring Potential

stretched or compressed springs

Kinetic EnergyKinetic Energy

A physical expression for kinetic energy can be derived using the work-energy theorem Consider an object that has a net force (FNET) applied to it over a distance (d)

FNET

vi

FNET

vfChange in motion

WNET = FNET d = m a d But vf2 = vi

2 + 2 a d So.. a = ( vf2 - vi

2 ) / 2d

WNET = m ( vf2 - vi

2 ) d =

2 d

or.. KEf - KEi = KE

What is the net work done on a 10 kg cart that increases its speed from 4 m/s to 15 m/s? What’s the force needed if the speed change occurs in a distance of 5 m

WNET = KE = 1/2 m (vf2 - vi

2)

m = 10 kg vi = 4 m/s vf = 15 m/s d = 5 m WNET = ? FNET = ?

= 1/2 (10kg) ( (15m/s)2 - (4m/s)2) = 1045 Nm

FNET = WNET / d = (1045 Nm) / 5m = 209 N

1/2 m vf2 - 1/2 m vi

2

Gravitational Potential EnergyGravitational Potential Energy

A physical expression for gravitational potential energy (GPE) can be derived using the work-energy theorem

Consider an object that is lifted a certain height at constant speed in a constant gravitational field

H

F

WT

+

-

F = WT = mg

W = F d

W = mg H

and… d = H

Because doing work always changes energy from one form to another then….

GPE = mg H = mg (df - di)

A 50 kg pile driver falls from 5m to 1m. How much GPE does it lose??

GPE = mg H = mg (df - di)

m = 50 kg di = 5 m df = 1 m g = 10 N/kg GPE = ?

= (50kg)(10N/kg) (1m- 5m)

GPE = - 2000 Nm = - 2000 J Note: negative means GPE has decreased

Transformation of EnergyTransformation of Energy

A device that changes energy from one form to another is called a machine

A car engine changes chemical energy into kinetic (moving car), gravitational potential energy (if car drives up a hill), and thermal energy (engine gets hot - exhaust gasses)

Car Engine -

Work is done by expanding gasses in a car engine cylinder pushing on the piston which is free to move

Plants -

Plants are natural machines. Nuclear energy in the sun is converted into radiant (EM) energy which is changed into chemical energy in the plant

Work is done by molecular transport ( ionic pump) across the plant (or animal) cell

Conservation of EnergyConservation of Energy

Conservative forces keep energy within a system (I.e. gravity)

Energy cannot be created nor destroyed, only transferred from one form to another

Non-conservative forces transfer energy out of a system (I.e. friction)

Written as an expression… KEi + PEi + WNC = KEf + PEf

Consider a car with 320 000J of KE braking on the flat with a force of 8000 N over a distance of 30m. What is the final energy of the car?

KEi = 320 000J d = +30 m F = -8000N PEi = PEf = 0 KEf = ? E

ne

rgy C

ha

ng

e

(ma

gn

itud

e)

KE = KEf - KEi - 240 000J

- WNC = + 240 000JKEi + PEi + WNC = KEf + PEf

KEi + WNC = KEf

KEi + F d = KEf

320 000J + (-8000N) (30m) = KEf

80 000 J = KEf

Conservation of EnergyConservation of Energy

What is the speed of the 50 kg jumper at B, C and D? Assume that there is no friction

m = 50 kg g = 10 m/s2 KEA = 0J PEA = 50 000 J dA = 100m dB = dD = 60m dc = 30m WNC = 0J vB

= ? vC = ? vD = ?

En

erg

y Ch

an

ge

A to

B

(ma

gn

itud

e)

KE = + 20 000 J

GPE = - 20 000 J

KEi + PEi + WNC = KEf + PEf

PEA = KEf + PEf

m g dA = 1/2 m vf2 + m g df

g dA = 1/2 vf2 + g df

g dA - g df = 1/2 vf2

2g( dA - df) = vf

At B: vB = 2g( dA - dB)= 2(10m/s2) (100m - 60m)

= 28 m/s

At C: vC = 2g( dA - dC)= 2(10m/s2) (100m - 30m)

= 37 m/s = 28 m/s At D: same height as at B so same speed

Force-Displacement GraphsForce-Displacement Graphs- How much work is done by a person pulling the cart 15m?

0

2

4

6

8

10

12

14

0 3 6 9 12 15

distance (m)

forc

e (

N)

The work done is the AREA under the applied force vs. displacement graph where the applied force is the component in the direction of motion.

AREA (rectangle) = h x b = 12N x 15m = 180 Nm

- How much work is done to stretch a spring in a spring scale 10cm?

0

5

10

15

20

25

30

0 2 4 6 8 10

distance (cm)

forc

e (

N)

2

The work done is the AREA under the applied force vs. displacement graph

AREA (triangle) =( h x b) / 2 = (25N x 0.1m) / 2 = 1.25 Nm

Note: This is the same as Fav d

PowerPower

Power is the rate at which work is donePower (P) = Work / Time = W / t Units: Nm / s or J/s or Watts (W)

P = Fav d / t = Fav vav

James Watt (1783) wanted to standardize the measure of power using something that everyone was familiar with ….. the power output of a horse.

If a large draft horse can pull 150 lbs while walking at 2.5 mi/h determine how many Watts one “horsepower” represents.

1 lb = 4.448 N 1 m/s = 2.237 mi/h

P = Fav vav = (150 lb) (4.448 N/lb) (2.5 mi/h) (1 m/s / 2.237 mi/h) = 746 W

PowerPower

An engine is used to raise a 2000 lb load 200 m vertically up a mine shaft. If the load travels upwards at a constant speed of 3 m/s calculate:

Fav = 2000 lb v = 3 m/s d = 200 m

i) P = Fav vav = (2000 lb) (4.448 N/lb) (3 m/s) = 26 688 W

a) The power rating of the engine in i) Watts and ii) Horsepower

Assume that the engine is 100% efficient (4.448 N = 1 lb)

= 30 000 W

ii) P (hp) = P (W) (1hp / 746 W) = 26 688 W (1hp / 746 W) = 36 hp = 40 hp

b) What is the power rating (hp) of the engine if it is only 70% efficient?

0.7 WIN = WOUT

0.7 WIN / t = WOUT / t

0.7 PIN = POUT

0.7 PIN = 36 W

Therefore… PIN = 36 W / 0.7 = 51 hp = 50 hp