Chapter 10

ENERGY, WORK, AND SIMPLE MACHINES

• Demonstrate a knowledge of the usefulness of simple machines.

• Differentiate between ideal and real machines in terms of efficiency.

• Analyze compound machines in terms of simple machines.

• Calculate efficiencies for simple and compound machines.

• Describe the relationship between work and energy.

• Calculate work.

• Calculate the work done by a variable force.

• Calculate the power used.

OBJECTIVESChapter 10 Section 1 Chapter 10 Section 2

ENERGY AND WORK10 - 1

Definition:

•Work is the transfer of energy by mechanical means.

•Work is done when a force is exerted on an object through a distance.

•Work is only done if the force is applied for a distance!

WORKW = Fd

Where:•W = work in joules (J)

•F = force in newtons (N)

•d = displacement in meters (m)

Note: A joule is equal to 1 Nm.

Energy Definition:

•The ability of an object to produce a change in itself or the world around it.

Kinetic Energy

•The energy resulting from motion.

•Represented by KE.

KE = ½ mv2

Where:

•KE = kinetic energy in joules (J)

•m = mass in kg

•v = velocity in m/s

KINETIC ENERGY

• The work-energy theorem states that when work is done on an object, the result is a change in kinetic energy.

• Determined by English physicist James Prescott Joule in the 1800’s.

W = ΔKEWhere:•W = work in joules (J)•ΔKE = change in kinetic energy

• KEf – KEi

• Measured in joules

WORK-ENERGY THEOREM

When a club head strikes a a 46-g golf ball, the ball picks up 43 J of kinetic energy. A constant force of 2300 N is applied to the ball while the club head and ball are in contact. Over what distance is the club head in contact with the ball?

Given:

m = 46 g

ΔKE = 43 J

F = 2300 N

d = ?

EXAMPLE - CALCULATING WORK Equations:

W = ΔKE

W = Fd

So, ΔKE = Fd

d = ΔKE / F

Calculations:

d = ΔKE / F

= 43 J / 2300 N

= 0.019 m

p. 261

# 2+3

HOMEWORK

xkcd.com

Remember:

•A force exerted in the direction of motion is given by W=Fd.

•A force exerted perpendicular to the direction of motion does no work.

So, how do you calculate work

when a constant force is exerted at

an angle?

So, how do you calculate work

when a constant force is exerted at

an angle?

CALCULATING WORK

CALCULATING WORK WHEN FORCE IS EXERTED AT AN ANGLE

1.Replace the force by it’s x and y components.

2.Determine which component of the force is in the direction of the object’s displacement.

3.Use this component of the force, in W = Fd, to calculate the work done.

Example:

How much work is done in pushing a tall box 15 m with a force of 4.0 x 102 N that is applied slightly upward at an angle of 10.0° from the horizontal?

Calculating Work – Force at an Angle

10.0° incline10.0° incline

15 m

400 N400 N

Given:F = 400 Nd = 15 m (in x-dir) W = ?

*Need F in x-dir.

Eqn & Calculations:W = FdF = F in x-dir = F cos θW = (F cos θ)dW = (400N)(cos 10)15mW = 5900 J

Fx

Fy

F

How can you calculate work when forces

change?

How can you calculate work when forces

change?

• Draw a force vs. displacement graph.

• The area under the curve is equal to the work done on the object.

FINDING WORK WHEN FORCES CHANGE

How do you calculate work when there is more than

one force?

How do you calculate work when there is more than

one force?

If several forces act on an object:

1. calculate the work done by each force

2. then add them together

* use components for forces that are at an angle

CALCULATING WORK

The work-energy theorem relates the net work done on the system to its change in

energy.

p. 262 #6-8HOMEWORK

xkcd.com

• Power is the rate at which the external force changes the energy of the system.

• The rate at which work is done.

• Measured in watts (W)• 1 W = 1 J/s

• Can also be measured in kilowatts (kW).

Calculating Power

Units:Power in watts (W) Work in Joules (J)Time in seconds (s)

POWER

A net force of 2800 N accelerates a 1250-kg vehicle for 8.0 s. The vehicle travels 80.0 m during this time. What power output does this represent?

Given:F = 2800 Nm = 1250 kgt = 8.0 sd = 80.0 mP = ?

EXAMPLE - CALCULATING POWEREquations:P = W / tW = Fdso:P = Fd / t

Calculations:P = Fd / tP = (2800 N)(80.0 m) / 8.0 s P = 28000 W or 28 kW

• pg. 264

• # 9-13

• This assignment will be checked before the end of class.

• QUIZ on Chapter 10 Section 1 TOMORROW!

• QUIZ topics include:

• Work

• Energy

• Work-Energy Theorem

• Power

CLASSWORK

MACHINES10 - 2

1.Lever

2.Wheel and Axle

3.Pulley

4.Inclined Plane

5.Wedge

6.Screw

WHAT ARE THE SIX SIMPLE MACHINES?

LEVER

WHEEL AND AXLE

PULLEY

INCLINED PLANE

WEDGE

SCREW

A machine is a device that

eases the load by achieving

one or more of the functions at

the left.

• transferring a force from one place to another

• changing either the magnitude of a force

• changing the direction of a force

• changing the distance or speed of a force

BENEFITS OF MACHINES

Effort Force (Fe)

•The force exerted by a person on a machine.

•Also called input force.

Resistance Force (Fr)

•The force exerted by the machine.

•Also called output force.

Mechanical Advantage

•The mechanical advantage of a machine is the ratio of the resistance force to the effort force.

MECHANICAL ADVANTAGE

• The resulting force from the machine is less than the applied force.

• So, the machine changes the distance or direction of the force.

• Ex. fishing pole

MECHANICAL ADVANTAGEMA > 1

• The resulting force from the machine is the same as the applied force.

• So, the machine changes the distance or direction of the force.

• Ex. pulley

• The resulting force from the machine is more than the applied force.

• So, the machine increases the force applied by the person.

• Ex. crowbar

MA = 1MA < 1

IDEAL MECHANICAL ADVANTAGE• In an ideal machine, all energy put in would be transferred out, so

the input work would be equal to the output work.

• Therefore, for an ideal machine, the mechanical advantage is equal to the displacement of the effort force divided by the displacement of the load – this is called ideal mechanical advantage (IMA).

EFFICIENCY

COMPOUND MACHINES