The Manga Guide to Physics

178 Chapter 4 Energy

Laboratory

The Relationship Between Work and Kinetic Energy

Let’s examine how we can derive an equation that expresses the relation-

ship between work and kinetic energy. Suppose we continue to impose

force F on a cart in motion, in a direction parallel to that cart’s velocity.

That cart has mass m and starts with an initial, uniform velocity of v.

That means an additional force is imposed on the object in motion.

At this time, the following is true:

work done on the object = Fd

Also, since we’ve represented the final velocity as v

2

, we can represent the

change in the object’s kinetic energy as the following:

change in kinetic energy = ½ mv

2

2

− ½ mv

1

2

And since we already know that the change in kinetic energy is equal to

the work done on the object, we can express the following relationship:

½ mv

2

2

− ½ mv

1

2

= Fd

Aha.

Initial velocity v

1

Final velocity v

2

Distance d, the distance that a force is applied

Force F

The Relationship Betwn Work and Kinetic Energy 179

We can also derive this equation another way. Since F is defined as con-

stant, the cart is experiencing uniform acceleration. Therefore, if we rep-

resent the cart’s acceleration with a, we know that the following must be

true:

v

2

2

− v

1

2

= 2ad

(Why is this so? See expression w on page 85.) To get closer to our original

expression, we’ll substitute using Newton’s second law:

F = ma, or rearranged just a little, a =

F

—

m

And we’ll get the following:

v

2

2

− v

1

2

=

2Fd

—

m

Then if you simply multiply both sides by ½, you’re there!

½ mv

2

2

− ½ mv

1

2

= Fd

I can get it right if I calculate very carefully.

Braking Distance

andSpeed

Using what we

know about the

relationship betwn

kinetic energy and

work, let’s consider

a car’s braking

distance.

What do you

mean, exactly?

We, I gue it’s

not just for cars.

It’s the distance that

any object in motion

requires to stop,

given a certain

force in the

oosite direction.

Given that we know a

change in kinetic energy

is equal to the work

performed, we know that

the foowing must be true

of bringing an object in

motion to rest:

½ ma × spd

2

= force of the brakes ×

distance the brakes are alied

½ mv

2

= F

brakes

× d

If we reaange

the equation, we

can solve for the

braking distance!

d =

½ mv

2

—

F

brakes

This equation means that

the greater the ma

(m) and the spd (v) of

the vehicle become, the

greater the required

distance to brake (d).

And the larger the

force of the brakes

( F

brakes

), the shorter the

distance required to

come to a complete

stop.

But we’ve

multiplied

the spd by

itself!?

That means that the

braking distance (d ) is

proportional to the

spd raised to the

second power.

The braking

distance is

proportional to

the spd raised

to the second

power.

When the initial

spd is doubled...

does that mean the

braking distance is

quadrupled?

Spd

braking distance

Aha, that is exceent

insight into their

relationship. It is

dangerous to aume

that the braking

distance is linearly

proportional to

acar’s spd.

Yeah,

definitely.

The braking distance

is in fact quadrupled

when your spd is

doubled.

For a bicycle, it’s

not such a big deal,

but for automobiles,

it can have serious

consequences.

For example, suose a

car is traveling at 40 km/h,

and its braking distance

is 10 m. If this same caris

traveling at 120 km/h,

or at a velocity thr

times higher, what is the

brakingdistance?

Um...

since the spd is

thr times higher, we

just have to square

that. So 3 × 3 = 9 times

greater, or

10 m × 9 = 90 m.