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Ch. 8: Conservation of EnergyPowerPoint Presentation

Ch. 8: Conservation of Energy

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Ch. 8: Conservation of Energy

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- Kinetic Energy: K = (½)mv2
- Associated with movement of members of a system

- Potential Energy
- Determined by the configuration of the system(location of the masses in space).
- Gravitational PE: Ug = mgy
- Elastic PE(ideal spring): Ue = (½)kx2

- Internal Energy
- Related to the temperature of the system

- Nonisolated systems
- Energy can cross the system boundary in a variety of ways
- Total energy of the system changes

- Isolated systems
- Energy does not cross the boundary of the system
- The total energy of the system is a constant
(CONSERVED!)

- Work– transfers energy by applying a force & causing a displacement of the point of application of the force
- Mechanical Waves– allows a disturbance to propagate through a medium
- Heat– is driven by a temperature difference between two regions in space

- Matter Transfer– matter physically crosses the boundary of the system, carrying energy with it
- Electrical Transmission– transfer by electric current
- Electromagnetic Radiation– energy is transferred by electromagnetic waves

a) Work b) Mechanical Waves c) Heat

d) Matter Transfer e) Electrical Transmission

f) Electromagnetic Radiation

Conservation of Energy

- TOTAL Energy is conserved
“Total”means the sum of all possible kinds of energy.

“Conserved”means that it remains constant in any process. In other words, Total Energy can be neither created nor destroyed, but only can be transformed from one form to another or transferred across a system boundary.

If the total amount of energy in a system changes, it can only be due to the fact that energy has crossed the boundary of the system by some method of energy transfer

The total change in the energy of a system = the

total energy transferred across a system boundary.

DEsystem = ST

Esystem= total energy of the system

T = energy transferred across the system boundary

- Established symbols: Twork = W & Theat = Q
- Others just use subscripts

A full expansion of the above equation gives:

D K + D U + DEint = W + Q + TMW + TMT + TET + TER

Isolated System

- For an isolated system,ΔEmech = 0
- Remember Emech = K + U
- This is conservation of energy for an isolated system with no nonconservative forces acting

- If nonconservative forces are acting, some energy is transformed into internal energy
- Conservation of Energy becomes: DEsystem = 0
Esystem is all kinetic, potential, & internal energies

The most general statement of the isolated system model

- For an isolated system, the changes in energy can be written out and rearranged
Kf + Uf = Ki + Ui

This applies only to a system in which only conservative forces act!

- Calculate the speed of the ball at a distance y above the ground
- Use energy
- System is isolated so the only force is gravitational which is conservative
- So, we can use conservation of mechanical energy!

- Conservation of Mechanical Energy
Kf + Ugf = Ki + Ugi

Ki = 0, the ball is dropped

- Solve for vf:
The equation for vfis consistent with the results obtained from kinematics

- Solve for vf:

Example 8.2 – Grand Entrance

- An actor, mass mactor = 65 kg, in a play is to “fly” down to stage during performance. Harness attached by steel cable, over 2 frictionless pulleys, to sandbag, mass mbag = 130 kg, as in figure. Need length R = 3 m of cable between nearest pulley & actor so pulley can be hidden behind stage. For this to work, sandbag can never lift above floor as actor swings to floor. Let initial angle cable makes with vertical beθ. Calculate the maximum value θcan have such that sandbag lifts off floor.

Free Body Diagrams

Step 1: To find actor’s speed at bottom, let yi = initial height above floor & use

Conservationof Mechanical Energy

Ki + Ui = Kf + Uf

or 0 + mactorgyi = (½)mactor(vf)2 + 0(1)

mass cancels. From diagram,

yi = R(1 – cos θ)

So, (1) becomes:(vf)2 = 2gR(1 – cosθ) (2)

Step 2: Use N’s 2nd Law for actor at bottom of path (T = cable tension).

Actor Sandbag

at bottom

Actor:∑Fy = T – mactorg = mactor[(vf)2/R]

or T = mactorg + mactor[(vf)2/R] (3)

Step 3: Want sandbag to not move. N’s 2nd Law for sandbag:

∑Fy = T – mbagg = 0 or T = mbagg (4)

Combine (2), (3), (4): mbagg = mactorg + mactor[2g(1 – cosθ)].

Solve for θ: cosθ = [(3mactor - mbag)/(2 mactor)] = 0.5 or, θ = 60°