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Kinetic Energy Kinetic energy is the energy of motion. By definition, kinetic energy is given by: K = ½ m v 2 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 v 2 , so doubling the speed quadruples kinetic energy, and tripling the speed makes it nine times greater. V V 2 then K.E. 4 K.E. 2

Mechanical energy

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A very important P.P. for the 9th grade students

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Page 1: Mechanical energy

Kinetic EnergyKinetic energy is the energy of motion. By definition,

kinetic energy is given by:

K = ½ m v 2

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 v 2, 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.

Page 2: Mechanical energy

Energy Units

The formula for kinetic energy, K = ½ m v 2, shows that its units are:

kg · (m/s)2 = kg · m 2 / s

2 = (kg · m / s 2

) 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.

Page 3: Mechanical energy

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!

Page 4: Mechanical energy

Gravitational Potential EnergyObjects 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 = m g 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

Page 5: Mechanical energy

SI Potential Energy Units

From the equation U = m g h 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.

Page 6: Mechanical 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

Page 7: Mechanical energy

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.

Page 8: Mechanical energy

Conservation of EnergyOne 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

Page 9: Mechanical energy

Example: SlideConsider 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   and is let go, what is its speed at the bottom?

Example: The SlideConsider 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” sothe 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

Page 10: Mechanical energy

WorkThe 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

Page 11: Mechanical energy

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.

Page 12: Mechanical energy

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

Page 13: Mechanical energy

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.

Page 14: Mechanical energy

Net WorkThe 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