Work & Energy AP/Honors Physics

Mr. Velazquez

What is “Work”?

When a force is applied to an object and that object moves

a distance as a result, work is being done

The amount of work done on an object by a constant force

(constant both in magnitude and direction) is defined as the

magnitude of the displacement times the magnitude of the

component of the force parallel to the displacement.

Work is calculated by multiplying two magnitudes (scalars),

which means it is a scalar quantity and not a vector.

The dimensions of work are Newtons (force) times meters

(distance), or 𝑵 ⋅ 𝒎—this combination of units is called the

joule (pronounced like “school”, abbrv. J)

Equations for Work

𝑊 = 𝐹∥𝑑

If the force is acting perfectly parallel to the direction of

movement, we simply multiply the magnitude of the

force by the distance.

𝑊 = 𝐹𝑑 cos 𝜃

If the force is acting at an

angle, we can multiply the

force by the cosine of the angle to obtain the component

that is parallel to the direction

of displacement.

Work Eqn 2

Work Eqn 1

Work Done by a Constant

Force A man pulls his daughter in a wagon behind him by applying a

constant force of 50 N at an angle of 30° above the horizontal as

shown. If the wagon travels a horizontal distance of 30 m, how

much work is done by the man?

𝑊 = 𝐹𝑑 cos 𝜃

𝑊 = 50 N 30 m cos 30° 𝑾 = 𝟏𝟑𝟎𝟎 𝐉

Work Done by Gravity

A 30.0-kg cannon ball falls straight

down to the ground from a height of

95.0 meters. What is the amount of work done on the cannon ball by

gravity?

𝑊 = 𝐹𝑑 cos 𝜃 = 𝐹𝑑 = 𝑚𝑔ℎ

𝑊 = 30.0 kg 9.807 ms2 95.0 m

𝑾 = 𝟐𝟕 𝟗𝟒𝟗 𝐉 = 𝟐𝟕. 𝟗 𝐤𝐉

𝐹𝑊

95.0 m In this example, the height is equal to the distance

the force is applied, or ℎ = 𝑑, and the force is

supplied by gravity in the form of weight, or 𝐹 =𝑚𝑔. The gravitational force acts in the same

direction of movement (downward), so the cosine

will cancel (because cos 0° = 1).

Positive and Negative Work

Suppose, in our previous example, the 30-kg

cannon ball is fired straight upward from the

ground and reaches a maximum height of 95.0

m. What is the work done on the cannonball

by gravity in this case?

95.0 m

𝐹𝑊

The applied force is opposite the direction of

displacement, so the magnitude of the force

will be negative, while the distance is positive:

𝑊 = 𝐹𝑑

𝑊 = −294 𝑁 95.0 𝑚

𝑾 = −𝟐𝟕 𝟗𝟒𝟗 𝐉 = −𝟐𝟕. 𝟗 𝐤𝐉

𝐹 = − 30.0 𝑘𝑔 9.807 𝑚𝑠2 = −294 𝑁

The same value as before, only negative!

Work Done by Gravity

A box weighing 40.0 kg is pulled down

a 15.0 m ramp which is inclined at an

angle of 35°. What is the amount of

work done on the box by gravity?

𝑊 = 𝐹𝑊𝑑 cos 𝜃

𝑊 = 392 N 15.0 m cos 55° 𝑾 = 𝟑𝟑𝟕𝟑 𝐉 = 𝟑. 𝟑𝟕 𝐤𝐉

𝐹𝑊

35°

𝜃

We need the angle 𝜃, which is

complimentary to 35°

𝜃 = 90° − 35° 𝜃 = 55°

We also need the magnitude

of the weight.

𝐹𝑊 = 40.0 kg 9.807 ms2

𝐹𝑊 = 392 N

Work Done by Gravity 𝐹𝑁

35°

𝜃 In the same example,

what is the amount of

work done on the box by

the normal force?

Since the normal force 𝐹𝑁 acts perpendicular to the

direction of movement, it has no component of its force in

the direction of motion. The cosine of the angle of

application is therefore zero, which means the work done by

the normal force is zero.

This will be true for any force that acts perpendicular to the

direction of movement.

Work Done by a Constant

Force (Graphical)

Work Done by a Variable

Force (Spring)

If the force is not constant, we

can find the work done by

finding the area underneath the

force vs. position graph.

To the left is the force required

to stretch a spring a length 𝑥

from equilibrium. We know how

to calculate the magnitude of

this force (𝐹 = 𝑘𝑥). Therefore the

area, under this graph will be a

triangle:

𝑊 =1

2𝑘𝑥 𝑥 =

𝟏

𝟐𝒌𝒙𝟐

Work Done by a Variable

Force (Other)

Exit Ticket, Part 1

1. A tourist pulls his suitcase forward, applying a force of

60.0 N at an angle of 50° above the horizontal. If he

pulls the suitcase a total of 100 m horizontally, how

much work is done on the suitcase by the tourist?

2. In order to come to a complete stop in 18.5 meters, a

particular car’s brakes must exert a braking force of 1150 N in the opposite direction. What is the amount of

work done on the car by the brakes?

𝑊 = 𝐹𝑑 cos 𝜃

Energy

What is Energy?

Work and energy are very closely linked

They can be considered two types of the same quantity,

even having the same units, joules

An object with energy can do work on another object

The total energy in a system will never change; it can

only be converted from one form to another

This energy can take many forms. We will examine two

specific types (for now):

Kinetic energy refers to the energy of motion

Potential energy is associated with forces that depend on

the position or configuration of an object(s) relative to the

surroundings

Kinetic Energy

Imagine an apple heading toward the ground with an

initial speed 𝑣1 and a final speed 𝑣2. It falls with

constant acceleration 𝑎. The gravitational force acting

on the apple produces a constant acceleration:

Applying the kinematic equation for velocity and

distance:

Substituting for 𝑎:

𝐹 = 𝑚𝑎 → 𝒂 =𝑭

𝒎

𝑣22 = 𝑣1

2 + 2𝑎𝑑 → 𝟐𝒂𝒅 = 𝒗𝟐𝟐 − 𝒗𝟏

𝟐

2𝐹

𝑚𝑑 = 𝑣2

2 − 𝑣12

𝐹𝑑 =1

2𝑚 𝑣2

2 − 𝑣12

𝐹𝑑 =1

2𝑚𝑣2

2 −1

2𝑚𝑣1

2 Work Change in

energy

Newton’s 2nd Law

Kinematic Eqn. 5

Kinetic Energy This quantity,

1

2𝑚𝑣2 is called the kinetic energy (abbrv. 𝐾 or

𝐾𝐸), and has a special significance in physics.

𝑊𝑡𝑜𝑡𝑎𝑙 = ∆𝐾 =1

2𝑚𝑣2

2 −1

2𝑚𝑣1

2

We can now define work as the total change in kinetic energy. This is known as the work-energy theorem:

𝐾 =1

2𝑚𝑣2

Source Approximate Kinetic

Energy (joules)

Jet aircraft at 500 mph 109

Car at 60 mph 106

Home-run baseball 103

Human at walking speed 50

Housefly in flight 10−3

Work-Energy Theorem

Kinetic Energy Eqn

Kinetic Energy

A truck moving at 16.0 m/s has a kinetic energy

of 154 kJ. What is the mass of the truck?

𝐾 =1

2𝑚𝑣2

𝑚 =2𝐾

𝑣2

𝑚 =2 154 × 103 𝐽

16.0 𝑚 𝑠 2

𝒎 = 𝟏𝟐𝟎𝟎 𝒌𝒈

Kinetic Energy Eqn

Kinetic Energy A boy pushes a 6.3-kg sled across the ice by exerting a

force of 13.0 N at an angle of 27.0° below the horizontal. Assuming it starts from rest, what is the work done by the boy and the final velocity of the sled after it moves 2.50 m?

Work done by boy:

𝑊 = 𝐹𝑑 cos 𝜃

𝑊 = 13.0 𝑁 2.50 𝑚 cos−27.0° 𝑾 = 𝟐𝟗. 𝟎 𝑱

Final speed of sled:

𝑊𝑡𝑜𝑡𝑎𝑙 = ∆𝐾 =1

2𝑚𝑣2

2 −1

2𝑚𝑣1

2

𝑊𝑡𝑜𝑡𝑎𝑙 =1

2𝑚𝑣2

2 → 𝑣2 =2𝑊𝑡𝑜𝑡𝑎𝑙

𝑚=

2 29.0 𝐽

6.3 𝑘𝑔

𝒗𝟐 = 𝟑. 𝟎𝟑 𝒎/𝒔

Potential Energy An object can have energy based on its position, connections, or

configuration. The object in question has the potential to either

do work or have work done on it. This is called potential energy.

Usually abbreviated 𝑼 or 𝑮𝑷𝑬.

Potential energy belongs to a system, not to a single object. The

type of potential energy depends on the type of system.

Gravitational Potential Energy is the product of the object’s weight,

and it’s height above the ground.

Elastic Potential Energy is the energy stored in stretched or

compressed springs or other elastic materials.

Others

Electrostatic Potential Energy: batteries and electrical sources

Magnetic Potential Energy: (what else?) magnets and electromagnets

Chemical Potential Energy: stored and released in chemical reactions

Gravitational Potential Energy

A brick held high in the air and a roller coaster at the peak of

its track both have gravitational potential energy (𝑼𝑮 or

GPE)—their height above the ground implies that gravity has

the potential to do work.

The higher the object is, the more potential energy it has.

Potential energy here would defined as the potential work

done by gravity, therefore:

𝑈𝐺 = 𝑚𝑔ℎ

𝑊𝐺 = ∆𝑈𝐺 = 𝑚𝑔ℎ1 −𝑚𝑔ℎ2

= −𝑚𝑔∆ℎ

Gravitational

Potential Energy

Work Done by Gravity = −𝑚𝑔(ℎ2 − ℎ1)

Gravitational Potential Energy

A coconut of mass 1.5 kg fall

from a palm tree 23.0 m high

onto a rooftop 17.0 m high.

What is the total work done

on the coconut by gravity

during its fall?

𝑊𝐺 = ∆𝑈𝐺 = −𝑚𝑔 ℎ2 − ℎ1

𝑊𝐺 = − 1.50 𝑘𝑔 9.807 𝑚𝑠2 (17.0 𝑚 − 23.0 𝑚)

𝑊𝐺 = 88.3 𝐽

Elastic Potential Energy

When a spring or some other elastic material is stretched or

compressed, an object connected to this system is said to have elastic potential energy (𝑼𝒆𝒍 or EPE).

The elastic potential energy is equal to the work done in

stretching the spring a length 𝑥 from its equilibrium.

𝑈𝑒𝑙 =1

2𝑘𝑥2

Elastic Potential Energy

Conservation of Mechanical Energy

The sum of the potential energy plus the kinetic energy in a system is called the total mechanical energy of the system.

If only certain forces are acting on a system, the total mechanical energy 𝐸 will neither increase nor decrease. This is the principle of conservation of mechanical energy.

These forces are called conservative forces; the total amount of work done by these forces does not depend on the path of motion.

∆𝐾 + ∆𝑈 = 0

Conservative

Forces

Nonconservative

Forces

Gravitational Friction

Elastic Fluid (air) resistance

Electric Tension in cord

Motor or rocket

propulsion

Push or pull by person

𝐾1 + 𝑈1 = 𝐾2 + 𝑈2

𝐸 = 𝐾 + 𝑈 Total Mechanical Energy

Conservation of Mechanical Energy (Conservative Forces Only)

𝐸1 = 𝐸2 = constant

Conservation of Mechanical Energy

(Nonconservative Forces)

If nonconservative forces are acting on a system, the total energy

at the end of the scenario will be less than the energy at the start.

The net change in energy will equal the work done by all

nonconservative forces.

∆𝐾 + ∆𝑈 = 𝑊𝑁𝐶 Conservative

Forces

Nonconservative

Forces

Gravitational Friction

Elastic Fluid (air) resistance

Electric Tension in cord

Motor or rocket

propulsion

Push or pull by person

𝐾1 + 𝑈1 +𝑊𝑁𝐶 = 𝐾2 + 𝑈2

Conservation of

Mechanical Energy

(Nonconservative Forces)

𝐸1 +𝑊𝑁𝐶 = 𝐸2

Conservation of Mechanical Energy

Conservation of Mechanical Energy

A roller coaster starts at a height of 14.0 m with an initial

velocity 𝑣1 and after its first dip, it ends up at a height of 4.0 m

with a velocity of 15.2 m/s. Assuming no friction or other

nonconservative forces are acting in this system, calculate 𝑣1.

𝐾1 + 𝑈1 = 𝐾2 + 𝑈2

1

2𝑚𝑣1

2 +𝑚𝑔ℎ1 =1

2𝑚𝑣2

2 +𝑚𝑔ℎ2

𝑚1

2𝑣1

2 + 𝑔ℎ1 = 𝑚1

2𝑣2

2 + 𝑔ℎ2

1

2𝑣1

2 + 𝑔ℎ1 =1

2𝑣2

2 + 𝑔ℎ2

1

2𝑣1

2 + 9.807𝑚

𝑠2(14.0 𝑚) =

1

215.2

𝑚

𝑠

2

+ 9.807𝑚

𝑠2(4.0 𝑚)

𝒗𝟏 = 𝟓. 𝟗𝟏 𝒎 𝒔

Conservation of Mechanical Energy

A 24.5-kg box slides across a frictionless surface with velocity 𝑣 and connects with a spring (𝑘 = 650 𝑁 𝑚 ). The length of the spring is then compressed by 0.75 m before the spring begins pushing the box back in the direction it came. What was the velocity 𝑣 of the box?

𝐾1 + 𝑈1 = 𝐾2 + 𝑈2

𝐾1 = 𝑈2

1

2𝑚𝑣2 =

1

2𝑘𝑥2

(24.5 𝑘𝑔)𝑣2 = 650 𝑁 𝑚 0.75 𝑚 2

𝒗 = 𝟑. 𝟖𝟔 𝒎/𝒔

Exit Ticket, Part 2

A spring with a force constant of 1225 𝑁 𝑚 is compressed a length of 1.15 m by a 17.5-kg box which is held in place, as shown below. When the box is released, it will slide over the surface and up a ramp (both the surface and ramp are assumed frictionless). Use the principle of conservation of mechanical energy to find:

(a) The speed of the box when it leaves the spring.

(b) The maximum height the box achieves after it reaches the ramp.

𝐾 =1

2𝑚𝑣2

𝑈𝐺 = 𝑚𝑔ℎ

𝑈𝑒𝑙 =1

2𝑘𝑥2

ℎ