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Kinetic Energy
Kinetic energy is the energy of motion. By definition,
kinetic energy is given by:
K = ½ mv2
The equation shows that . . .
. . . the more kinetic energy it’s got.
• the more mass a body has
• or the faster it’s moving
K is proportional to v2, so doubling the speed quadruples
kinetic energy, and tripling the speed makes it nine times greater.
V  V2
then K.E.  4 K.E.
V  V2
then K.E.  4 K.E.
Energy Units
The formula for kinetic energy, K = ½ m v2, shows that its units are:
kg·(m/s)2 = kg·m2 / s2 = (kg·m / s2)m = N ·m = J
So the SI unit for kinetic energy is the Joule, just as it is for work.
The Joule is the SI unit for all types of energy.
One common non-SI unit for energy is the calorie. 1 cal = 4.186 J.
A calorie is the amount of energy needed to raise the temperature of
1 gram of water 1 C.
A food calorie is really a kilocalorie. 1 Kcal = 1000 cal = 4186 J.
Another common energy unit is the British thermal unit, BTU, which
the energy needed to raise a pound of water 1 F. 1 BTU = 1055 J.
Kinetic Energy Example
A 55 kg toy sailboat is cruising at
3 m/s. What is its kinetic
energy?
This is a simple plug and chug
problem:
K = 0.5 (55) (3)2 = 247.5 J
Note: Kinetic energy (along with
every other type of energy) is a
scalar, not a vector!
Gravitational Potential Energy
Objects high above the ground have energy by virtue of their height.
This is potential energy (the gravitational type). If allowed to fall, the
energy of such an object can be converted into other forms like kinetic
energy, heat, and sound. Gravitational potential energy is given by:
U = mg h
The equation shows that . . .
. . . the more gravitational potential energy it’s got.
• the more mass a body has
• or the stronger the gravitational field it’s in
• or the higher up it is
SI Potential Energy Units
From the equation U = mgh the units of gravitational
potential energy must be:
kg·(m/s2) ·m = (kg·m/s2) ·m = N·m = J
This shows the SI unit for potential energy is the Joule, as it is for
work and all other types of energy.
Reference point for G.P.E. is arbitrary
Gravitational potential energy depends on an object’s height, but
how is the height measured? It could be measured from the floor,
from ground level, from sea level, etc. It doesn’t matter what we
choose as a reference point (the place where the potential energy is
zero) so long as we are consistent.
Example1: A 20kg mountain goat is perched atop of a 220 m
mountain ledge. How much gravitational potential
energy does it have?
G.P.E.= mgh = (20) (10) (220) = 44000J = 44KJ
This is how much energy the goat has with respect to the ground
below. It would be different if we had chosen a different reference
point.
continued on next slide
Reference point for G.P.E. (cont.)
Example2: The amount of g.p.e. the watermelon
has depends on our reference point. It can be
positive, negative, or zero.
10 N
D
C
B
A
6 m
3 m
8 m
Reference
Point
Potential
Energy
A 110 J
B 30 J
C 0
D 60 J
Note: the weight of the object
is given here, not the mass.
Conservation of Energy
One of the most important principles in all of science is “conservation of
energy”. It is also known as the first law of thermodynamics. It states that
“energy can change forms, but it cannot be created nor destroyed”. This
means that “the total energy in a system before any energy
transformation must be equal to the total energy afterwards”.
For example, suppose a mass is dropped from some height.
The gravitational potential energy it had originally is not
destroyed. Rather it is converted into kinetic energy
(ignoring friction with air). The initial total energy is given
by Ei = G.P.E.= mgh. The final total energy is given by
Ef = K.E. = ½ mv 2.
Conservation of energy demands that Ei = Ef (ignoring
energy loss).
Therefore, in general: Ei = Ef +|Elost| , (where Ei >Ef )
For this example in general:“mgh = ½ m v 2 + Elost ”.
m
Before
v=0
After
h=0
v
m
Example: The Slide
Consider the slide below, and assume that its quite frictionless. If
you prefer, we could be more adventurous and imagine that this is a
roller coaster. Don't get too carried away, we got to keep our minds
on the physics here 
If the object shown starts out at rest at a height (h) and is let go, what
is its speed at the bottom?
Solution:
The initial state of the slide:
The initial height of the object is “h”
and its initial kinetic energy is zero.
Therefore the initial mechanical
energy is M.E.i =K.E.+G.P.E.=mgh
The final state:
The final value of height is zero
So G.P.E.=0, and its final speed “v” so
the final energy is M.E.f =K.E.+G.P.E
=1/2 mv2
Energy is conserved  M.E.i + M.E.f
 Mgh =1/2 mv2
 v2
= 2gh
Work
The simplest definition for the amount of work a force does on an object
is magnitude of the force times the distance over which it’s applied:
W = F * D
This formula applies when:
• the force acting on the object is constant
• The force is moving with object
• the force is in the same direction as the displacement of the object
F
D
Work Example
50 N
10 m
The work this applied force does is independent of the presence of any
other forces, such as friction. It’s also independent of the mass.
A 50N horizontal force is applied to a 15kg box over a distance of 10m.
The amount of work this force does is
W = 50 N · 10 m = 500 N · m
The SI unit of work is the (Newton × Meter). There is a shortcut for
this unit called the Joule, J. 1 Joule = 1 Newton × meter, so we can say
that the this applied force did 500 J of work on the crate.
Negative Work
Distance moved = 7 m to the rightFriction = 20 N to the left
Or Friction = -20N
A force that acts opposite to the direction of motion of an object does
negative work. Suppose the box slides across the floor until friction
brings it to a stop. The displacement (motion) is to the right, but the
force of friction is to the left. Therefore, the amount of work friction
does is -140 J, meaning that this work causes the object to lose 140J of
energy.
Friction doesn’t always do negative work. When you walk, for
example, the friction force is in the same direction as your motion, so it
does positive work in this case.
v
When zero work is done
7 m
N
mg
As the box slides horizontally, the normal force and weight do no
work at all, because they are perpendicular to the displacement. If the
box was moving vertically, such as in an elevator, then then each force
would be doing work. Moving up in an elevator the normal force
would do positive work, and the weight would do negative work.
Another case when zero work is done is when the displacement is
zero. Think about a weight lifter holding a 20 Kg barbell over her
head. Even though the force applied is 20 Kg, and work was done in
getting over her head, no work is done just holding it over her head.
Net Work
The net work done on an object is the sum of all the work done on it
by the individual forces acting on it. Work is a scalar, so we can
simply add work up.
Example: In the following figure, the applied force does +200 J of
work; friction does -80 J of work; and the normal force and weight
do zero work.
So, Wnet = 200 J - 80 J + 0 + 0 = 120 J
F = 50 N
4 mFr = 20 N
N
mg
Note that (Fnet )×(distance) = (50-20 N) (4 m) = 120 J.
Therefore, Wnet = Fnet × d

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Mechanical energy

  • 1. Kinetic Energy Kinetic energy is the energy of motion. By definition, kinetic energy is given by: K = ½ mv2 The equation shows that . . . . . . the more kinetic energy it’s got. • the more mass a body has • or the faster it’s moving K is proportional to v2, so doubling the speed quadruples kinetic energy, and tripling the speed makes it nine times greater. V  V2 then K.E.  4 K.E. V  V2 then K.E.  4 K.E.
  • 2. Energy Units The formula for kinetic energy, K = ½ m v2, shows that its units are: kg·(m/s)2 = kg·m2 / s2 = (kg·m / s2)m = N ·m = J So the SI unit for kinetic energy is the Joule, just as it is for work. The Joule is the SI unit for all types of energy. One common non-SI unit for energy is the calorie. 1 cal = 4.186 J. A calorie is the amount of energy needed to raise the temperature of 1 gram of water 1 C. A food calorie is really a kilocalorie. 1 Kcal = 1000 cal = 4186 J. Another common energy unit is the British thermal unit, BTU, which the energy needed to raise a pound of water 1 F. 1 BTU = 1055 J.
  • 3. Kinetic Energy Example A 55 kg toy sailboat is cruising at 3 m/s. What is its kinetic energy? This is a simple plug and chug problem: K = 0.5 (55) (3)2 = 247.5 J Note: Kinetic energy (along with every other type of energy) is a scalar, not a vector!
  • 4. Gravitational Potential Energy Objects high above the ground have energy by virtue of their height. This is potential energy (the gravitational type). If allowed to fall, the energy of such an object can be converted into other forms like kinetic energy, heat, and sound. Gravitational potential energy is given by: U = mg h The equation shows that . . . . . . the more gravitational potential energy it’s got. • the more mass a body has • or the stronger the gravitational field it’s in • or the higher up it is
  • 5. SI Potential Energy Units From the equation U = mgh the units of gravitational potential energy must be: kg·(m/s2) ·m = (kg·m/s2) ·m = N·m = J This shows the SI unit for potential energy is the Joule, as it is for work and all other types of energy.
  • 6. Reference point for G.P.E. is arbitrary Gravitational potential energy depends on an object’s height, but how is the height measured? It could be measured from the floor, from ground level, from sea level, etc. It doesn’t matter what we choose as a reference point (the place where the potential energy is zero) so long as we are consistent. Example1: A 20kg mountain goat is perched atop of a 220 m mountain ledge. How much gravitational potential energy does it have? G.P.E.= mgh = (20) (10) (220) = 44000J = 44KJ This is how much energy the goat has with respect to the ground below. It would be different if we had chosen a different reference point. continued on next slide
  • 7. Reference point for G.P.E. (cont.) Example2: The amount of g.p.e. the watermelon has depends on our reference point. It can be positive, negative, or zero. 10 N D C B A 6 m 3 m 8 m Reference Point Potential Energy A 110 J B 30 J C 0 D 60 J Note: the weight of the object is given here, not the mass.
  • 8. Conservation of Energy One of the most important principles in all of science is “conservation of energy”. It is also known as the first law of thermodynamics. It states that “energy can change forms, but it cannot be created nor destroyed”. This means that “the total energy in a system before any energy transformation must be equal to the total energy afterwards”. For example, suppose a mass is dropped from some height. The gravitational potential energy it had originally is not destroyed. Rather it is converted into kinetic energy (ignoring friction with air). The initial total energy is given by Ei = G.P.E.= mgh. The final total energy is given by Ef = K.E. = ½ mv 2. Conservation of energy demands that Ei = Ef (ignoring energy loss). Therefore, in general: Ei = Ef +|Elost| , (where Ei >Ef ) For this example in general:“mgh = ½ m v 2 + Elost ”. m Before v=0 After h=0 v m
  • 9. Example: The Slide Consider the slide below, and assume that its quite frictionless. If you prefer, we could be more adventurous and imagine that this is a roller coaster. Don't get too carried away, we got to keep our minds on the physics here  If the object shown starts out at rest at a height (h) and is let go, what is its speed at the bottom? Solution: The initial state of the slide: The initial height of the object is “h” and its initial kinetic energy is zero. Therefore the initial mechanical energy is M.E.i =K.E.+G.P.E.=mgh The final state: The final value of height is zero So G.P.E.=0, and its final speed “v” so the final energy is M.E.f =K.E.+G.P.E =1/2 mv2 Energy is conserved  M.E.i + M.E.f  Mgh =1/2 mv2  v2 = 2gh
  • 10. Work The simplest definition for the amount of work a force does on an object is magnitude of the force times the distance over which it’s applied: W = F * D This formula applies when: • the force acting on the object is constant • The force is moving with object • the force is in the same direction as the displacement of the object F D
  • 11. Work Example 50 N 10 m The work this applied force does is independent of the presence of any other forces, such as friction. It’s also independent of the mass. A 50N horizontal force is applied to a 15kg box over a distance of 10m. The amount of work this force does is W = 50 N · 10 m = 500 N · m The SI unit of work is the (Newton × Meter). There is a shortcut for this unit called the Joule, J. 1 Joule = 1 Newton × meter, so we can say that the this applied force did 500 J of work on the crate.
  • 12. Negative Work Distance moved = 7 m to the rightFriction = 20 N to the left Or Friction = -20N A force that acts opposite to the direction of motion of an object does negative work. Suppose the box slides across the floor until friction brings it to a stop. The displacement (motion) is to the right, but the force of friction is to the left. Therefore, the amount of work friction does is -140 J, meaning that this work causes the object to lose 140J of energy. Friction doesn’t always do negative work. When you walk, for example, the friction force is in the same direction as your motion, so it does positive work in this case. v
  • 13. When zero work is done 7 m N mg As the box slides horizontally, the normal force and weight do no work at all, because they are perpendicular to the displacement. If the box was moving vertically, such as in an elevator, then then each force would be doing work. Moving up in an elevator the normal force would do positive work, and the weight would do negative work. Another case when zero work is done is when the displacement is zero. Think about a weight lifter holding a 20 Kg barbell over her head. Even though the force applied is 20 Kg, and work was done in getting over her head, no work is done just holding it over her head.
  • 14. Net Work The net work done on an object is the sum of all the work done on it by the individual forces acting on it. Work is a scalar, so we can simply add work up. Example: In the following figure, the applied force does +200 J of work; friction does -80 J of work; and the normal force and weight do zero work. So, Wnet = 200 J - 80 J + 0 + 0 = 120 J F = 50 N 4 mFr = 20 N N mg Note that (Fnet )×(distance) = (50-20 N) (4 m) = 120 J. Therefore, Wnet = Fnet × d