Conservation of Total Energy

Physics 11

Review: Law of conservation of mechanical energy:

Prime (‘) used to represent conditions after process has completed All units = Joules Don’t have to use all 3, depends on situation

But we are not including/ignore things like friction, air resistance, etc.

Ek + EP +Es = Ek’ + Ep

’ +Es’

Systems and forces:

System: any object or group of objects can be defined as a system

Once the system is defined, forces are classified as internal or external: Internal force: force exerted by

something in the system on anything in the system

External force: force exerted by any object that is not part of the system

Open-system:

Both matter and energy can freely cross from the system to the surroundings and back.

Ex: an open test tube, pot of boiling water

Closed-System:

Energy can cross the boundary, but matter cannot.

Ex: a sealed test tube, pot of water with a cover

Isolated-System: Neither matter nor

energy can cross between the system and the surroundings.

Ex: The universe there are no

surroundings to exchange matter or energy with (as far as we know!)

Energy Conservation Closed systems (mechanical energy

systems) are straight forward

Open systems are not, however, energy is still conserved.

FI EE

ntOnEnvionmeFI

LostFI

WEE

EEE

Energy Conservation

This is the same as considering the definition of work, but now it is the work done by the environment on the system, so the sign is reversed.

ntByEnvionmeFI

IF

WEE

EEEW

Change in Energy

Both equations are correct, but you have to remember which way the energy is transferring in each.

Energy lost to friction is a positive value

Work done by friction is negative

ntOnEnvionmeFI WEE

ntByEnvionmeFI WEE

Strategies:

These questions have various forms and can incorporate a lot of difference formulas we’ve used (especially W and energy)

Generally try to find the energy before and after, then subtract to get energy lost

Or if given the energy lost, use to figure out energy before or after as necessary

Example:

A bullet (1.75g) passes through a glass door with a velocity of 91.2 m/s and exits the glass with a velocity of 13.9 m/s. How much work did the friction do to slow

down the bullet? If the glass is 2.97cm thick, what was the

average force exerted by the glass on the bullet?

Example 1:

m=0.00175kg, vi=91.2m/s, vf=13.9 m/s Can assume horizontal movement….Ek only!

Ek(i) = ½ mv2 (Ek bullet before)

= ½(0.00175)(91.2)2

= 7.3 J

=Ek(f) = ½ mv2 (Ek bullet after) = ½(0.00175)(13.9)2

= 0.17 JWork done by friction = 7.3-0.17 = 7.13 J

Example 1(continued..):

Work done by friction = 7.3-0.17 = 7.13 J

W = F· d

7.13 = F (0.0297)

240 N = F

Try it :

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