physics chapter 10 section 1 work, energy, and power

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Work, Energy, and Power

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Work, Energy, and Power

Work is done on a system when a force is applied through a displacement.

Work is measured in joules.

One joule of work is done when a force of 1N acts on a system over a displacement of 1m .

Work

Work

Consider a force is exerted on an object while

the object moves a certain distance.

There is a net force, so the object is

accelerated, a =𝑓

π‘š, and its velocity changes.

Work done by a constant force

Recall from the equations of motion 𝑣𝑓2 = 𝑣𝑖

2 + 2π‘Žπ‘‘

This can be written as𝑣𝑓2 βˆ’ 𝑣𝑖

2 = 2π‘Žπ‘‘

If you replace (a) with (𝑓

π‘š) you obtain :

𝑣𝑓2 βˆ’ 𝑣𝑖

2 = 2(𝑓

π‘š)𝑑 multiplying both sides by

π‘š

212π‘šπ‘£π‘“

2 βˆ’ 12π‘šπ‘£π‘–

2 = 𝑓𝑑

𝐾𝐸𝑓 βˆ’ 𝐾𝐸𝑖 = 𝑓𝑑

π‘Š = 𝑓𝑑

Work done by a constant force

Work done by a constant force exerted at an angle

𝐹π‘₯ = 𝐹 cos πœƒ = 125𝑁 cos 25Β° = 113 𝑁.𝐹π‘₯ = 𝐹 sin πœƒ = 125𝑁 sin 25Β° = βˆ’52.8 𝑁.

The negative sign shows that the force is downward.

Because the displacement is in the x direction, only the x-component does work. The y-component does no work.

The work you do when you exert a force on a system at an angle to the direction of motion is equal to the component of the direction of the displacement multiplied by the displacement.

Work is equal to the product of the magnitude of the force and magnitude of displacement times the cosine of the angle between them.

π‘Š = 𝐹𝑑 cos πœƒ

W : work

F : force

d : displacement

πœƒ : the angle between the two directions of both force and displacement

Work

When several forces are exerted on a system, calculate the work done by each force, and then add the results.

Suppose you are pushing a box on a friction less surface while your friend is trying to prevent you from moving it, as shown in the following figure.

Work done by many forces

Work done by many forces

What are the forces are acting on the box?

You are exerting a force to the right and your friend is exerting a force to the left.

Earth’s gravity exerts a downward force, and the ground exerts an upward normal force.

Work done by many forces

How much work is done on the box? The upward and downward forces are perpendicular πœƒ = 90Β° to the direction of motion and do no work. For these forces, πœƒ = 90Β°, which makes cos πœƒ = 0, and thus π‘Š = 0.The force you exert is in the direction of the displacement, so the work you do is W = πΉπ‘œπ‘› π‘π‘œπ‘₯ 𝑏𝑦 π‘¦π‘œπ‘’ Γ— 𝑑Your friend exerts a force in the direction opposite the displacement πœƒ = 180Β° . Because cos 180Β° = βˆ’1, your friend does negative work. W = βˆ’πΉπ‘œπ‘› π‘π‘œπ‘₯ 𝑏𝑦 π‘“π‘Ÿπ‘–π‘’π‘›π‘‘ Γ— 𝑑The total work done on the box would be

W = (πΉπ‘œπ‘› π‘π‘œπ‘₯ 𝑏𝑦 π‘¦π‘œπ‘’Γ— 𝑑) βˆ’(πΉπ‘œπ‘› π‘π‘œπ‘₯ 𝑏𝑦 π‘“π‘Ÿπ‘–π‘’π‘›π‘‘ Γ— 𝑑)

Work done by many forces

In the last example, the force changed, but we could determine the done in each segment. But what if the force changes in a more complicated way?

A graph of force versus displacement lets you determine the work done by a force.

This graphical method can be used to solve problems in which the force is changing.

Finding work done when forces change

Finding work done when forces change

The left panel of this figure shows the work done by a constant force of 20.0

N that is exerted to lift an object 1.50 m.

The work is done by this force is represented by π‘Š = 𝐹𝑑 =

20𝑁 1.5π‘š = 30𝐽

Notice that the shaded area under the left graph is also equal to

20𝑁 1.5π‘š , π‘œπ‘Ÿ 30𝐽

The area under a force-displacement graph is equal to the force done by

that force.

Finding work done when forces change

The right panel of this figure shows the force exerted by a

spring, which varies linearly from 0 to 20 N as it

compressed 1.5 m. the work done by the force that

compressed the spring is the area under the graph, which

is the area of a triangle, π΄π‘Ÿπ‘’π‘Žβˆ† =1

2π‘π‘Žπ‘ π‘’ π‘Žπ‘™π‘‘π‘–π‘‘π‘’π‘‘π‘’ , or

π‘Š = 1

220𝑁 1.5π‘š = 15 𝐽

Finding work done when forces change

Use the problem-solving strategies below when you solve

Energy is the ability of a system to produce a change in it self or the world around it.

Energy

Work-Energy theorem states that work done on a system is equal to the change in the system’s energy.

𝑀 = βˆ†πΈ

Energy is measured in joules.

Work-Energy Theorem

Through the process of doing work, energy can move between the external world and the system.

The direction of energy transfer can be either way.

if the external world does work on a system, then Wis positive and the energy of the system increases.

If a system does work on the external world, then Wis negative and the energy of the system decreases.

Work is the transfer of energy that occurs when a force is applied through a displacement.

Work-Energy Theorem

Kinetic energy (KE) is the energy associated with motion.

Translational kinetic energy (πΎπΈπ‘‘π‘Ÿπ‘Žπ‘›π‘ ) is the energy due to changing position.

πΎπΈπ‘‘π‘Ÿπ‘Žπ‘›π‘  =12π‘šπ‘£2

A system’s translational kinetic energy is equal to 1

2

times the system’s mass multiplied by the system’s speed squared.

Changing kinetic energy

Suppose you had a stack of books to move from the floor to a shelf.

You could lift the entire stack at once, or you could move the books one at a time.

How would the amount of work compare between the two cases? Since the total force applied and the displacement are the same in both cases, the work is the same.

Power

The time needed is different. Recall that work causes a change in energy. The rate at which energy is transformed is power.

Power is equal to the change in energy divided by the time required for the change.

𝑃 =βˆ†πΈ

𝑑

Power

When work causes the change in energy, power is equal to the work done divided by the time taken to do the work.

𝑃 =π‘Š

𝑑 Power is measured in watts (W).

One watt is 1 J of energy transformed in 1 s.

1,000 W = 1 KW 1,000,000 W = 1 MW

Power

You might have noticed in Example problem 3 that when the force has a component 𝐹π‘₯ in the same

direction as the displacement, 𝑃 =𝐹π‘₯𝑑

𝑑.

However, because 𝑣 =𝑑

𝑑, power also can be calculated

using 𝑃 = 𝐹𝑣

Power and speed

Power and speed