Download Conservation of Energy

Law of Conservation of Energy:
Energy can not be created or
destroyed.
Remember E=mc2
That is to say, energy
and mass are
interchangeable.
Actually, It’s
conservation of
mass - energy
Although energy cannot be destroyed,
it can go places where we can never recover
it.
Potential
Kinetic
Chemical
Energy
Heat
RIP
Electrical
Sound
And many more…
Example: A diver weighing 750. N (mass =
77.0 kg) drops from a board 10.0 m above
the surface of the pool of water.
A) Use the conservation of
mechanical energy to find her
speed at a point 5.00 m
above the water surface.
10 m
B) Find the speed of the diver
just before she strikes the
water.
Conservation of Mechanical Energy
Although energy may be changed into a
different form, the final value will be the
same as the initial value (energy is not
lost).
W net = KEf – KEi
W net = Wc = PEi – PEf
So, PEi – PEf = KEf – KEi
Or, KEi + PEi = KEf + PEf
½ mvi2 + mgyi = ½ mvf2 + mgyf
Example: A diver weighing 750. N (mass = 77.0 kg) drops
from a board 10.0 m above the surface of the pool of water.
A) Use the conservation of mechanical energy to find
her speed at a point 5 m above the water surface.
B) Find the speed of the diver just before she strikes the water.
yf = 5 m
½ mvi2 + mgyi = ½ mvf2 + mgyf
½(77)(0)+(77)(9.8)(10)=½(77)(vf2 )+(77)(9.8)(5)
Vf = 9.86 m/s
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Example: A diver weighing 750 N (mass = 77.0 kg) drops from
a board 10 m above the surface of the pool of water.
A) Use the conservation of mechanical energy to find her speed
at a point 5 m above the water surface.
B) Find the speed of the diver just before she strikes
the water.
Example: A sled and its rider together weigh 800. N.
They move down a frictionless hill through a vertical
distance of 10.0 m. Use conservation of energy to
find the speed of the sled-rider system at the bottom
of the hill, assuming the rider pushes off with an
initial speed of 5.00 m/s.
yf=0
½ mvi2 + mgyi = ½ mvf2 + mgyf
½ mvi2 + mgyi = ½ mvf2 + mgyf
½ vi2 + gyi = ½ vf2 + gyf
½(5)2+ (9.8)(10)=½(vf2)+ (9.8)(0)
½(77)(0)+(77)(9.8)(10)=½ (77)(vf2)+(77)(9.8)(0)
vf = 14.9 m/s
10 m
Vf = 13.9 m/s
Kinetic vs. Potential Energy
Kinetic vs. Potential Energy
Calculate the KE and PE energy at each of the 5 spots on the
ski run. Assume the skier starts from rest. (m = 51 kg)
Calculate the KE and PE energy at each of the 5 spots on
the roller coaster. Assume the skater starts from rest.
Conservative / Nonconservative
Forces
Conservative / Nonconservative
Forces
A
A
force is CONSERVATIVE if the work
it does on an object between 2 points is
independent of the path the object takes
between the 2 points. (The work done
depends only on the initial and final
positions.) (example: Potential Energy)
PE = mgy
W c = -(PEf - Pei)
force is NONCONSERVATIVE if the
work it does on an object moving
between 2 points depends on the path
taken. (i.e. sliding friction)
 **Potential
energy functions cannot be
defined for nonconservative forces.
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Conservation of Energy with
Nonconservative Forces.
A
15 kg kid, initially at rest, slides down
a 8.0 m high slide. Ideally, what his is
final velocity at the bottom of the slide?
 KEi + PEi = KEf + Pef
 0 + mgy = 1/2 mv2 + 0
 (15kg)(9.8m/s2)(8.0m) = 1/2 (15kg)v2
 vf = 12.5m/s ~13m/s
KEi + PEi + Energyadded = KEf + PEf + Energylost
Example:
Work of someone
pushing
Example:
Work done by
friction
Note: Usually work done by friction is negative (because
it is in the opposite direction). But since we are
ADDING BACK IN the work done by friction it is
positive.
Potential Energy Stored in a Spring
PEs = ½ kx2
where k = spring constant
x = distance spring is compressed or stretched
(KE + PEg + PEs)i = (KE + PEg + PEs)f
In reality, the final velocity of the
kid is only 10 m/s. Where did
the extra energy go?
Friction is an nonconservative force which used
up some of the initial potential energy. To
account for friction, we need to add back in
the work done by (or energy lost by) friction.
How much work was done by
friction as the kid slid down the
slide?
KEi + PEi + Energyadded = KEf + PEf + Energylost
0 + mgy + 0 = 1 /2mvf2 + 0 + Wfriction
(15kg)(9.8m/s2)(8.0m) = 1/2 (15kg)(10.m/s)2 + Wf
1176 J = 750 J +Wf
Wf = 426 J ~ 430J
Example: A block of mass 0.500 kg rests on a horizontal,
frictionless surface. The block is pressed lightly against a
spring, having a spring constant k=80.0 N/m. The spring is
compressed a distance of 2.00 cm and released. Find the speed
of the block at the instant it loses contact with the spring at the
x = 0 position.
(KE + PEg + PEs)i = (KE + PEg + PEs)f
0.500 kg
(PEs)i = (KE)f
½kxi2 = ½mvf2
½(80)(0.02)2 = ½(0.5) vf2
v = 0.253 m/s
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Example: The launching mechanism of a toy gun consists of a
spring of unknown spring constant. By compressing the spring
a distance of 0.120 m, the gun is able to launch a 20.0-g
projectile to a maximum height of 20.0 m when fired vertically
from rest. Determine the value of the spring constant.
Example:
A 5.0 kg block is given an initial velocity of 3.0
m/s at the top of a hill. Calculate the distance the
spring will be compressed when it stops the block.
(KE + PEg + PEs)i = (KE + PEg + PEs)f
0
0
0
0
(PEs)i = (PEg)f
½kxi2 = mgy
½(k)(0.12)2 = (0.02)(9.8)(20)
k = 544 N/m
Friction µ = 0.20
20.m
10.m
Spring k=100.
30.o
Work done by Friction
Or Energy dissipated through friction
N = mgcos30o
PEtop + KEtop =Ffr d + ½ kx2
Mgy + ½ mv2 = µN d + ½ kx2
mgy + ½ mv2 = µmg(cos30o) d + ½ kx2
(5)(9.8)(10) + ½ (5)(3)2 = (.2)(5)(9.8)cos30o(20) +½(100)x2
50x2 = 342.5
X = 2.62 m
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