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Energy
Chapters 7 & 8
10/3/2011
Potential + Kinetic = Total Energy
Where
P.E = mgh and K.E = ½ mV2and E = W = F x d
Where Total Energy is conserved
Hooke’s Law
• Mathematically:
– Fspring = -kx
• Where k = the spring constant in N/m
• And x = (x – xequilibrium) in meters
• Definition:
– The force required to stretch/compress a spring is
proportional to x (the amount of stretch/compression)
• A “stiffer” spring will cause an increased spring
constant, where a “looser” spring will decrease
the spring constant
W = Force X distance for a constant
force. For a changing force W =
• W=
Fdx = kxdx = k xdx
• W = kx2/2 from xi to xf = potential energy
• Since work = Force X distance, to hold a spring
in place it requires no work, just force. Work is
only done when the spring is moving!
Delta x oscillates in a sin wave as time
goes on when a spring is released
For purely frictionless mechanical
things that travel in straight lines PE &
KE are constant so :
delta PE (x) = - delta KE (x)
Conservative System
• It doesn’t matter how (what path) you move
in—as you move from higher potentials you
gain kinetic energy and as you move from
lower potentials you lose kinetic energy.
• Friction makes a system not-conservative. You
must factor in heat in this case
– KE + PE + Heat = Total energy
– (where KE & PE are mechanical energy)
Rate of Change of Energy
• Delta Energy / delta time = power
• OR Work / delta time (for strictly mechanical
energy) = power
• Power units is J/s = watt
• Examples of power/energy relationships:
– Heat takes a lot of energy, so it requires more
watts (AC unit) whereas sound takes next to no
energy and uses very little power (the stereo)
Examples
• *The ball to the right
is falling with an
increasing speed. As
its speed increases, its
kinetic energy grows.
Where is this energy
coming from, since
the laws of physics say
that energy cannot be
created or destroyed?
• The increasing kinetic
energy is coming from
the initial gravitational
potential energy and
the potential energy as
it falls toward the
earth. As shown to the
right, the ball begins
with a potential energy
of 600 J and 0 J of
kinetic energy, but
ends with 0 J of
Potential energy and
600 J of Kinetic energy.
No energy is created
or destroyed
If a cannon sits at the top of a 50 m tall cliff and
fires a cannonball into the air over the edge at
an angle of 30 degrees and the initial velocity of
the ball is 20m/s, what is the speed just before
the cannon hits the ground?
20 m/s
30o
------------------I
50m
I
I
I
I
Solving without Energy
Vi=20m/s Yi=50m Yf=0m θ= 300
• Yf= yi + Vyit + ½ayt2
• 0 = 50m + 20sin30o + ½ (-9.8m/s2) t2
• 0 = 50m + 10m/s – 4.9m/s2 t2
• Quadratic Results: -2.3s & 4.4s
• a t + Vxi = vyf
• Vyf = (-9.8m/s2)(4.4s) + 20cos300
• Vyf = -32.8m/s
• Vf = √(Vfx2 + Vfy2)
• Vf = √(17.32 +32.92)
• Vf = 37.1m/s
Seems like a lot of math…
Solving Using Energy
•
•
•
•
•
•
ΣEi = ΣEf
GPEi + Kei = Kef
mgh + ½mvi2 = ½mvf2
Vf = √(2gh + vi2)
Vf = √[(2)(9.8m/s2)(50m) + 20m/s]
Vf = 37.1m/s
Easier right? Be careful, though
this only works when the question
does not ask for a direction
A 10 kg block is sliding down a ramp inclined at 200 from the
ground. The block and ramp have a coefficient of kinetic friction
of 0.2. The top of the ramp is 2 m above the ground. How much
energy was lost to friction as the block slides down the ramp?
What is the final speed of the block when it reaches the bottom?