Chapter 3, Energy and
Conservation Laws
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Physics 2001, Professor Ray Merry |
Conservation Laws:
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What does it mean for a physical quantity to be conserved? |
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It means: |
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The amount of that thing does not
change. |
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E.G. Conservation of mass, means the
amount of mass in a system stays constant. |
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Take the example of a burning
cigarette. |
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Cigarette before burning weighs 10 g. |
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After burning the butt plus the ash
weighs 5 g. |
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What happened? Is mass conserved? |
Burning Cigarette
What is the Closed System
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Note for all the conservation laws we
must include all the parts of the system! |
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For example, not considering the air in
the cigarette example led to an error. |
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This is the secret to good scientific
analysis. What is the total
environment or system? |
Linear momentum, p:
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The mass of an object times its
velocity. |
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p = mv Note since v is a
vector so is p! |
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Example, m = 5 kg, v = 10 m/s what is
p? |
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p = 5 x 10 kg m/s = 50 kg m/s |
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Note units are kg m/s or mass x vel.
Units. |
Newton’s Second Law w
Impulse
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F= D(mv)/Dt is another way of stating Newton’s second law, in terms of |
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D(mv),
change in momentum and |
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Dt time
for the change to occur. |
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We can also state this as |
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F Dt = D(mv) Where F Dt is the impulse of a force. |
Impulse Diagram-Ball
Hitting Floor
Impulse Problem
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Suppose a ball of .1 kg is in contact
with a bat for .2 s, and the force the bat exerts is 5 n. What would the speed of the ball be? |
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F Dt = D(mv) since original velocity of the ball is 0, D(mv) = m x final vel.(vf) |
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5n x .2s = .1kg x v final |
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Vf = 1 ns/.1kg = 10 m/s |
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(n=kg m/s/s nxs = kg m/s n/kg = m/s2) |
Law of Conservation of
Linear Momentum:
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The total linear momentum of an
isolated system is constant. |
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Remember this is a vector relationship |
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If all velocities are in the same
direction it is simple, if not we must use vectors to solve |
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total mv before = total mv after |
1 D Momentum Problem
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A 200lb boy moving at 10 ft/sec
hits 2 50 kg midgets at rest.
Both go off together, but how fast ? |
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Momentum before = momentum after |
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Mom.before =200 kg x 10m/s = 2000 kgm/sec |
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Mass after = 50 +50 + 200 = 300 kg |
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After mom = 300 x V2; 2000 = 300
V2 |
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V2 = 2000/300 = 6.6 m/s |
Work
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The force that acts times the distance
moved in the direction of the force. |
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Work = F x d (Joules =Newtons x meters) |
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Work units are Joules, same as energy. |
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E.G. Force of 10 N moves 5 m |
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W = 10 N x 5 m = 50 Nm or 50 J |
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Note Force is a vector but work is a
scalar!c |
Force vs Work
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Note work = force x distance in
direction of force. |
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If Force is perpendicular to motion or
displacement no work is done! |
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Examples circular motion, moving
parallel to floor |
Gravitational Work
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When we lift a body we do work against
gravity, but when a body falls, gravity does work on it. If the distances are equal so are the
works. |
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Work = F x d |
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F = weight = W=mg |
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W=Wd = mgd |
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Example 10kg falls 5m, what is the
work? |
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W = 10kg x 9.8m/s/s x 5m =490J |
Energy (E)
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The measure of a system's capacity to
do work. |
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That which is transferred when work is
done. |
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Energy is measured in Joules, which are Newtons x meters. |
Define kinetic energy:
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Kinetic energy (KE) Energy due to
motion. Energy that an object has when it is moving |
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KE = ½ mv2 |
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Example, 20 kg mass moving at 15 m/s |
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KE = ½ x 20 kg x (15)2 =10 x
225 = 2250 J |
Potential Energy
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Potential energy (PE) is Energy due to
an object's position or configuration. |
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A stretched spring has potential
energy. |
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A raised object has gravitational
energy. |
Gravitational Potential
Energy
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Energy due to position, PE |
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Gravitational PE = mgd |
Other PE’s
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elastic potential energy |
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stretched out rubber band or spring. |
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internal energy |
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unburned fuel, battery |
PE Problem
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We can calculate the potential energy
of a given situation if we know the force. |
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Since work against gravity is weight x
height lifted. mgd, PE is the same,
mgd. |
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W = m g d. Lift 10 kg, 2 meters |
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W = 10kg x 9.8 m/s/s x 2 m |
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W = 196 J; PE stored = same, 196 J |
Work and Potential Energy
Law of Conservation of
Energy:
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Energy cannot be created or destroyed,
only converted from one form to another. |
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The total energy in an isolated system is constant. |
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Eo = Ef (original
energy = final energy) |
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In terms of PE and KE: |
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PE + KE = constant |
Conservation of Energy
Problem
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A man does a work against gravity of 98
Joules lifting an object. |
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What is its KE before it strikes if it
falls back down? |
Energy Conversion and
Conservation
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Many common devices convert energy from
one form to another. |
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For example your radio converts
chemical energy from a battery to electrical to sound energy. |
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Although the law of conservation of
energy holds, some of the energy ends up in forms we do not need or want, for
example as heat. |
Energy Conservation in
Society
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When we talk about energy conservation
in society we mean |
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Don’t waste energy |
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Renewable energy is energy that we can
recover soon, like wood, or solar energy |
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Fossil fuels like coal and oil take
millions of years to replenish |
Elastic collision
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Collision in which the total kinetic
energy of the colliding objects is the same before and after the collision. |
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I.E. KE is conserved. |
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½ mv2 before = ½ mv2 after |
Inelastic collision
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Collision in which the total kinetic
energy of the colliding bodies after the collision is not equal to the total
kinetic energy before. |
Torque (levers)
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Torque is the rotational analog of
force. |
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Torque = force times lever arm |
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Example teeter totter |
Inclined Plane
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Inclined plane increases distance moved
but decreases force needed. |
Power
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Power (P): The rate of doing work. |
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The rate at which energy is transferred
or transformed. |
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Work done divided by the time. Energy
transferred divided by the time. |
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Unit of power is the watt (joule/s) |
Power Problem
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20 Joules of work is done in 5 s. What
is the power? |
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P = W/t = 20J/5s = 4 watts |
Angular momentum
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The mass of an object times its speed
and times the radius of its path. |
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mvr |
Law of conservation of
angular momentum
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The total angular momentum of an
isolated system is constant. mvr=
constant |
Ice Skater Spinning
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When the skater brings their arms in,
since angular momentum is constant and r is decreasing, v must increase. |
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