Conservation of Energy The work-energy theorem says

EKUW If no external force does any work on an

object (or system), then the energy of the system does not change

The total energy is the same at beginning and end (and all times between)

The energy of an isolated system is conserved

Work Work is the transfer of energy to an object

by applying external force. Work is not a vector. It has a sign, but no

direction in space. The work done by a force, on an object,

is positive if the force is giving energy to the object.

depends on the size of the force and the displacement through which it is applied.

In 1-D, if force is constant:

extW F d

Work — Units Units of work or of energy Since

extW F d

Units of work = (units of force)*(unit of distance)

Units of work = N m also called a Joule

1 J 1 N m

Work—Lift Example

I lift a 2-kg weight up onto a 1.1 m-tall table. How much work do I do?

a) Suppose I lift straight up at constant speed.

b) Suppose slide up a 22o slope at constant speed.

extW F d

Lift Example 1a) Given d = 1.1m, what F do I need to

exert?Constant velocity No net force, so my

force to lift each is F = mg = (1kg)(9.8m/s2)Work done lifting each:(9.8 N)(1.1 m) 10.8 JextW F d

Note: F and d in same direction (both positive or both negative) make W positive.

Total work done lifting both: 21.6 J

Lift Example 2b) Given h = 1.1m, d=1.1/sin(22o) =

2.936 mwhat F do I need to exert? Sum of forces

in x-direction (along slope) =0, soF = mgsin() = (2kg)(9.8m/s2)sin(22o)

=7.342NWork done sliding:

(7.342 N)(2.936 m) 21.6 JextW F d

Total work done sliding: 21.6 J

Lift Example 3 I can’t lift the weight at constant

velocity, if it isn’t moving to begin with. Suppose I accelerate it at a rate of 2.1m/s2 for 0.25 s, then lift at constant speed, then slow it back down at a rate of 2.1 m/s2 for 0.25 s.

No extra work is done overall in speeding it up, then slowing it back down to its original speed (here zero).

Lift Example 3Part 1- speeding it up. F=ma F-mg=ma, soF= mg+ma = (2kg)(9.8m/s2)+(2kg)(2.1m/s2) = 23.8Nd=½ at2 =½ (2.1m/s2) (.25)2 = 0.065625mW1=(23.8N)(0.065625m)= 1.561875JPart 3 – slowing down F=ma F-mg=ma, soF= mg+ma = (2kg)(9.8m/s2)+(2kg)(-2.1m/s2) = 15.4Nd again= 0.065625mW3=(15.4N)(0.065625m)= 1.010625 JPart 2 – constant speed F=ma F-mg=0, soF= mg = (2kg)(9.8m/s2) = 19.6Nd = rest of the distance = 1.1m – 2(0.065625m) = 0.96875 mW2=(19.6N)(0.96875 m)= 18.9875 JTotal work = 1.561875J+ 1.010625 J+ 18.9875 J = 21.6J!

Coincidence? No! Total work done raising weight is 21.6 J, whether you lift straight up or up a

slope, at constant speed or not, …because the weights have gained 21.6 J

of energy from me, independent of how I go about giving it to them.

21.6 J is how much energy it takes to make this particular change (raising) in the state of the weights, no matter what.

Work on a SpringW = Fd

Only works if F is constant throughout displacement, Spring force is not constant F=-kx

Work on a Spring Calculus to the rescue: If we consider a

small displacement, dx, over which the force is _________________, then the small amount of work, dW, is

dW = Fdx To get the total work,

2212

21

i

x

x

f kxkxkxdxWf

i

Spring Potential Energy Once again, if the force does work on the

spring

Where does that energy go? Like with gravity, it is

2212

21

ifspring kxkxUW

Work in 2-D and 3-D

Two ways to calculate this: W=Fxdx+Fydy+Fzdz

using components of the vectors, or W=Fdcos()

where F and d are magnitudes, and = angle BETWEEN F and d NOT angle from x-axis

W F d

Conservation Example

ffii KUKU

A 35-g ball is placed on a compressed spring, which shoots it straight up. The spring has a spring constant of 220 N/m, and is initially compressed by 3.5 cm. Neglecting drag, how high does the ball go?Do we have an isolated system?

Use:

h = 0.393 m or 39 cm above where it started

Drag Drag is resistance to motion of an object

through a fluid If fluid is air, sometimes called air

resistance Drag with streamline, non-viscous flow

depends on: fluid density (), cross-sectional area of

object (A), speed of object relative to fluid (v), properties of object’s surface (C).

Cross-sectional area can be thought of as the area of the shadow the object would have, if lit from the direction of the passing fluid.

Nf ss max,

Drag Depends on:

density of fluid (), cross-sectional area of object presented to fluid (A), relative speed of object and fluid (v), properties of the object’s surface (C).

Nf ss max,

212D C Av

Direction: Always opposes relative motion of fluid and object

Note: This eqn doesn’t apply to viscous or turbulent flow

Projectiles and drag An object moving vertically does have

the same vertical motion as an object that is moving sideways too, even if vyi is same, if drag is not negligible.

Drag force has a vertical component that depends on speed, not just vy

Nf ss max,