172 Chapter 4 Energy

Laboratory

Work and the Conservation of Energy

Let’s consider a scenario in which we are lifting a heavy load to a certain

height. The simplest way to do this is to lift straight up. The following dia-

gram shows how it looks.

We are lifting a load with mass m to height h.

Let’s consider how much work we must do to lift the load to a height of h

by imposing a force equal to the force of gravity of the mass—that is, we’ll

impose a force upward equivalent to the force downward from gravity.

Assuming g for gravitational acceleration, we know that the force down-

ward is mg:

work upward = force of lifting × height h = mgh

Note that for simplicity’s sake, we won’t take into account friction or air

resistance in these examples. But this is a hard way to lift something so

heavy!

Hmm . . . maybe it’d be easier if we pushed the load up a ramp.

Yes, let’s consider the case of pushing the load up an incline.

Lifting force

h

mg

Work and the Conservation of Energy 173

Look at this diagram. The magnitude of the force needed to push the load

up this ramp (F) is equal to the component of the force of gravity parallel

to the ramp (PR). So, if the ramp has a length of d, the work required to

move the load to height h can be represented as:

work = Fd

Now, you know intuitively that F is smaller than mg, and d is larger than h.

That makes sense. Is that why it takes the same amount of work to push

the load up a ramp as it does to lift the load straight up?

Yes, indeed. Now let’s show why this works, mathematically. r ABC repre-

sents the ramp in the figure, and rPQR represents the composition of the

force mg. These two triangles are similar—this means that ∠CAB = ∠RPQ.

This also means that the proportion of their corresponding sides must be

the same, as well. Thus, the following must be true:

AB

—

AC

=

PQ

—

PR

Let’s make this a little less abstract. The line segment AB equals d (length

of ramp) and AC equals h (height). Similarly, the line segment PQ equals

mg (the force downward, due to gravity), while PR equals F (the force

applied to offset a portion of that force).

Lifting

force (F)

A

CB

R

P

Q

Component of

weight (PR)

d

Weight (mg)

h

174 Chapter 4 Energy

That means:

d

—

h

=

mg

—

F

Look, with just a little rearranging of this equation we get the following:

Fd = mgh

Therefore, the work to lift a load using a ramp must be equal to the work

to lift that load straight upward.

Also, please note that our results are the same, regardless of the angle

of the ramp. Given the conservation of energy, regardless of the lifting

route, the work done for lifting an object with mass m to height h is equal

to the following:

force required to balance gravity × height = mgh

So, whatever method you use to lift something, the amount of work you

do is the same.

To put it another way, your work increases the potential energy of the load

by mgh.

And I bet it works for negative work, too. That is, you’d see a decrease in

potential energy of mgh if you lower an object by mgh.

Yep, that’s right.

Work and Energy

Work isn’t

onlydone when

increasing or

decreasing

potential

energy.

Work can

alsoaect the

kinetic energy

ofan object!

You mean work is

also done when

we move an object

or bring a moving

object to a halt?

Are you sti

listening to me,

Ninomiya-san?

Yes, go on

with the

leon.

We.

Ahem.

While you impose a

force for a given

distance on an object

at rest, that object’s

kinetic energy

increases.

Imposing a

force on an

object

Generates

kinetic energy.

What’s

haening?

I fl

like I’m

shrinking...

I’m a

grown

up!

Aw!

How

cute!

What the...?

F

Squal

Whoa. This is true

forobjects in

motion as we. In

other words, the

kinetic energy of

an object increases

even more

if you impose

a force in the

direction of the

object’s motion.

For some

reason, you

remind me of a

pachinko ba.

Since energy is

conserved, we know

thefoowing:

work done on the object =

change in the object’s kinetic energy

This relationship

must hold true.

Ah, yes.

If the force we

impose on an

object is in the

direction of the

object’s

motion—

that is, when the

force and velocity

are parael—we

wi do positive

work.

176 Chapter 4 Energy

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