Ch. 8: Conservation of Energy. Energy Review. Kinetic Energy : K = ( ½) mv 2 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 : U g = mgy
a) Work b) Mechanical Waves c) Heat
d) Matter Transfer e) Electrical Transmission
f) Electromagnetic Radiation
Conservation of Energy
“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
A full expansion of the above equation gives:
D K + D U + DEint = W + Q + TMW + TMT + TET + TER
Esystem is all kinetic, potential, & internal energies
The most general statement of the isolated system model
Kf + Uf = Ki + Ui
This applies only to a system in which only conservative forces act!
Kf + Ugf = Ki + Ugi
Ki = 0, the ball is dropped
The equation for vfis consistent with the results obtained from kinematics
Example 8.2 – Grand Entrance
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:∑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°