When work is done by a completelyunbalanced forceit can set objects in motion…or increasetheir speed.

You push and it simply coasts away beyond your reach.

d

F

d

F

F

d

When an objectmoves in the

direction of an unbalanced push

it accelerates over the entire

distance d.

W = F d

produces the acceleration

a = F/m acceleratingthrough the distance din a time t: d = ½ at2

Work is done on a wagon, initially at rest, by pushing it with an unbalanced force F over a distance D. With whatfinal speed does it coast away?

Final speed? atvv 0

With what acceleration a does it build speed?

mFa /

But for how long does it accelerate?2

21 atd 22 atd

2/2 tad adt /2

ada

dav 2

2 Sofinally

W = F dBut there’s another way to think about this

F = ma d = at212

W = ma at2

So the amount of work done is also equal to:

12

W = ma2t212

atvv 0

Then, just recall that theacceleration builds up, over t, to a final velocity:

0, if starting from rest!

W = mv212

We consider this an expression of the energy possessed by a mass m

when moving with a velocity v

Moving objects carry energy

to scatter pins to knock down walls

In fact, energy is defined as the ability to perform work!

push snow aside

A piledriver uses a raised weight to drive piles into the ground.

Raised weights also possess energy…potential energy…the potential to fall and build kinetic energy.

Raising a mass against gravity is doing work, even if there is no resulting final velocity.

Work done in lifting a mass to a height h:

W = F daverage force

lifted withmg

W = mgh

In falling through a distance h a massbuilds speed to what final kinetic energy?

2

21 gth 2

21 atd

ght /2And in that time reaches

ghghgatv 2/2

)2(212

21 ghmmvKE

KE = mgh

So:

Weight

Weight

Normal Force

If tipped completelyof course the ramp

gives NO SUPPORT!

A horizontalsurface (if strong

enough to hold up!) can give full support to an object’s weight

What about any of thepositions in between?

Weight

Normal Force

Normal Force

The unbalanced (net) force is parallelto the inclined surface of the road!

Exactly the direction we know a carleft in neutral will start to coast!

We can “add”these forces

in a diagramby showingthe arrows

that representthem, flowing

from headinto tail from

one to the other(in either order

in fact!)

h d

Pushing this box up along the ramp is 1. more work2. less work3. exactly the same work

as simply lifting it straight up h.

x

d

WN

F

Wd

=F?

1. h 2. N 3. x

h d

Work pushing box along the ramp is

Work in simply lifting the box h

x

d

WN

F

Wd

=Fh

Force in lifting distance liftedW h

F d

FdWh

A man, his briefcase in hand as shown,walks down the corridor from elevator to office door. He does1. positive work on the briefcase.2. no work on the briefcase.3. negative work on the briefcase.

F

d

A man, his briefcase in hand as shown,walks down the corridor from elevator to office door. He does1. positive work on the briefcase.2. no work on the briefcase.3. negative work on the briefcase.

F

d

The force F does is not along (nor back against) the direction of motion. Instead F and d are perpendicular!The force of support neither raises nor lowers the briefcase (neither increasing or decreasing its potential energy). ItNeither speeds up nor slows down the briefcase (neitherincreasing nor decreasing the briefcases kinetic energy).

h d

x

d

WN

F

Wd

=Fh

The box may slip down from its position on the shelf by sliding back down the ramp or falling straight down to the floor.By the time it reaches the floor it will reach

1. a higher final speed by falling.2. a higher final speed by sliding.3. the same final speed by either route.

assumefrictionlessrollers on

ramp

h dd

WN

F

Wd

=Fh

Falling: 2

21 gth ght /2

ghghggtvv 2/200

Sliding: mFa / Wd

hF mg

d

h

d

hg

m

mg

d

ha

2

2

1t

d

hgd

hgdt /2 2

hg

d

d

hgv

22

hg2then

Some Answers

3. exactly the same workQuestion 1

This won’t be obvious until we work through

the next several questions!

1. hQuestion 2

The small triangle drawn to display the forces actingother box, and the large triangle formed by the ramp,wall and floor are similar. d is the hypotenuse of the large triangle, W the hypotenuse of the small one.

2. no work on the briefcase.Question 3

The force F does is not along (nor back against) the direction of motion. Instead F and d are perpendicular! The force of support neither raises nor lowers the briefcase (neither increasing or decreasing its potential energy). It neither speeds up nor slows down the briefcase (neither increasing nor decreasing the briefcases kinetic energy).

3. the same final speed by either route.Question 4

The small triangle drawn to display the forces actingother box, and the large triangle formed by the ramp,wall and floor are similar. d is the hypotenuse of the large triangle, W the hypotenuse of the small one.

But already this much should be clear: lifting straight up requires a large force.Pushing along the ramp a smaller force, but the push must be applied over amuch larger distance (the full length of the ramp). As you’ll see by the end ofthis lecture the proportions in the triangles guarantee the two compensate exactly!The large force time small distance = the smaller force times longer distance.