Download Section 10.1 Energy and Work

Mr. Borosky
Physics Section 10.1 Notes
Page 1 of 4
Chapter 10 Energy, Work, and Simple Machines
In this chapter you will:
Recognize that work and power describe how the external world
changes the energy of a system.
Relate force to work and explain how machines ease the load.
Sections
Section 10.1: Energy and Work
Section 10.2: Machines
Section 10.1 Energy and Work
Objectives
Describe the relationship between work and energy.
Calculate work.
Calculate the power used.
Read intro paragraph p. 257
Conserved Properties – properties that are the same before and after
an interaction. Examples are energy and momentum.
WORK AND ENERGY
Read Section.
Work – is the product of the force and the object’s displacement.
It is equal to the constant force exerted on an object in the
direction of motion times the object’s displacement. It is the
transfer of energy by Mechanical means. It is denoted by W. It is
measured in Joules.
W = Fd
Energy – the ability of an object to produce a change in itself or
in its surroundings.
Kinetic Energy – is equal to ½ times the mass of an object times the
speed of the object squared. It is denoted by KE. It is measured
in Joules.
KE = ½ mv2
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 10.1 Notes
Page 2 of 4
Work Energy Theorem – states that work is equal to the change in
Kinetic energy.
W = ΔKE
James Prescott Joule – physicist that established the relationship
between work done and the change in energy that results. The unit
of Energy is named after him.
Joule – unit of Work. One Joule equals 1 Newton meter which equals
1 kg*m2/s2 (1 J = 1 Nm = 1 kg*m2/s2).
If the external world does work on a system then “W” is positive and
the energy of the system increases.
If the system does work on the external world then “W” is negative
and the energy of the system decreases.
CALCULATING WORK
Read Section.
Work is done on an object only if it moves. If you hold an object
in place you do not do any work. Also if you move an object at
constant velocity at a constant height you do no work (because the
force is up and the motion is sideways not in the same direction of
the force).
Work is only done when the force and displacement are in the same
direction.
A force displacement graph can give you a picture of the work done.
The area under the curve of a force displacement graph represents
the work done.
If Force and Displacement are at right angles then W = 0.
Remember that you can replace a force by it’s component parts (the x
and y parts of the Vector Force).
The work you do when you exert a force at an angle to a motion is
equal to the component of force in the direction of the motion times
the distance moved.
The magnitude of the component of force F acting in the direction of
motion is found by multiplying the force F by the cosine of the
angle between F and the direction of motion.
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 10.1 Notes
Page 3 of 4
Work at an Angle – is equal to the product of the force and
displacement times the cosine of the angle between the force and the
direction of the displacement.
W = Fd cos 
The work done by Friction acts in the direction opposite that of
motion. So the work done by Friction is Negative.
Negative work reduces the Kinetic Energy of the system.
Do Example Problem 1 p. 261
W = Fd
W = ΔKE
W = 4.5(.15)
.675 J = ΔKE
W = .675 Joules
Do Practice Problems p. 261 # 1-3
Do Example Problem 2 p. 262
W = Fd cos 
W = 255(30) cos 25°
W = 7650(.90631)
W = 6,933.25 J
Do Practice Problems p. 262 # 2-8
Work is the area under the curve in a graph of Force versus
Displacement.
If several forces are exerted on a system, calculate the work done
by each force and then add the results.
POWER
Read Section.
Power – is the work done divided by the time taken to do the work.
It is denoted by “P”. It is measured in Watts.
Watt – the unit of Power.
It is equal to 1 Joule per second.
Power is often measured in Kilowatts since a watt is so small.
= 1000 watts.
1 KW
Physics Principals and Problems © 2005 Started 2006-2007 School Year
Mr. Borosky
Physics Section 10.1 Notes
Do Example Problem 3 p. 264
P = W / t = Fd / t
P = 12,000(9) / 15
P = 108,000 / 15
P = 7,200 Watts
or
7.2 KW
also
Page 4 of 4
7,200 J/s
Do Practice Problems p. 264 # 9-14
Since Displacement divided by time (d / t) = velocity we can find
Power using
P = Fv
Do 10.1 Section Review p. 265 # 15-23
Physics Principals and Problems © 2005 Started 2006-2007 School Year