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 Betwn 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

andSpeed

Using what we

know about the

relationship betwn

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

oosite direction.

Given that we know a

change in kinetic energy

is equal to the work

performed, we know that

the foowing must be true

of bringing an object in

motion to rest:

½ ma × spd

2

= force of the brakes ×

distance the brakes are alied

½ mv

2

= F

brakes

× d

If we reaange

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 spd (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 spd by

itself!?

That means that the

braking distance (d ) is

proportional to the

spd raised to the

second power.

The braking

distance is

proportional to

the spd raised

to the second

power.

When the initial

spd is doubled...

does that mean the

braking distance is

quadrupled?

Spd

braking distance

Aha, that is exceent

insight into their

relationship. It is

dangerous to aume

that the braking

distance is linearly

proportional to

acar’s spd.

Yeah,

definitely.

The braking distance

is in fact quadrupled

when your spd is

doubled.

For a bicycle, it’s

not such a big deal,

but for automobiles,

it can have serious

consequences.

For example, suose a

car is traveling at 40 km/h,

and its braking distance

is 10 m. If this same caris

traveling at 120 km/h,

or at a velocity thr

times higher, what is the

brakingdistance?

Um...

since the spd is

thr times higher, we

just have to square

that. So 3 × 3 = 9 times

greater, or

10 m × 9 = 90 m.

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