Monday, October 5, 1998

Chapter 5: SpringsChapter 6: Linear Momentum

Conservation of Momentum Impulse

A 600 kg elevator startsfrom rest and is pulledupward by a motor witha constant accelerationof 2 m/s2 for 3 seconds.What is the averagepower output of the motorduring this time period?

Fnet = ma = (600 kg)(2 m/s2)

Fnet = 1200 N = Fmotor - Fg = Fmotor - mg

1200 N = Fmotor - (600 kg)(10 m/s2) = Fmotor - 6000 N

Fmotor = 6000 N + 1200 N = 7200 N

Let’s first figure out theforce delivered by themotor...

A 600 kg elevator startsfrom rest and is pulledupward by a motor witha constant accelerationof 2 m/s2 for 3 seconds.What is the averagepower output of the motorduring this time period?

Now we need todetermine the workdone by the motor...

W = F s But we don’t know s, so….

s = s0 +v0t +0.5at2 = 0 + 0 + 0.5(2 m/s2)(3 s)2 = 9 m

W = (7200 N)(9 m) = 64800 J

PW

t

64800

321600

J

s W

In addition to the gravity, there are othermechanisms to store POTENTIAL ENERGY.One of them is...

Sir Robert Hooke unlocked the secretof the spring...

A spring resting in its natural state, with alength l exerts no horizontal force on anything!

However, if we compress or stretch the springby some amount x, then the spring is observedto exert a Force in the opposite direction.

Hook discovered this force could be modeledby the mathematical expression

F = - kxNotice that this force operates along a linear line!

l x

Which means that if we looked at the plotof Force versus compression/stretching x...

For

cex

Slope of this lineis -k, where k isthe spring constant.

For

ce

x

If we look at the work done by an appliedforce which compresses the spring througha distance (-x1)...

-x1

F1

W F s Fx F x 11

2 1 0( )

W k x x kx 1

2 1 11

22( )( )

Work done BY theexternal force ONthe spring.

This energy is stored in the spring...

PE kxspring 1

22Potential Energy of a spring is

So, for spring problems, we have a newTOTAL MECHANICAL ENERGY given by

KE PE PEg spring

And it is THIS quantity which will be conservedabsent other, outside forces.

Momentum & Collisions

The linear momentum of an object of mass mmoving with velocity v is defined as theproduct of the mass and the velocity:

p mv

p mv

Notice that momentum is a vector quantity,which means that it must be specified withboth a magnitude and direction.

Also notice that the direction of themomentum vector is necessarily parallelto the velocity vector.

p mv

[ ] [ ][ ] p m v

[ ]p kg (m / s)

[ ]p N s

OR The units suggesta relationship betweenforce and momentum.

p mv

What happens when we apply a forceto an object?

It accelerates.

Its velocity changes.

The force imparts momentum.

Its momentum changes.

p mv

By how much will the momentum change?

That depends upon the length of time overwhich the force is applied to the object.

v v a t 0

mv mv ma t 0

mv mv F t 0

mv mv F t 0

p p F t 0

p F t I Impulse

Change inmomentum

The impulse of a force on an object equals thechange in momentum of that object.

Notice that impulse is a vector quantity as well!