Download Elastic Potential Energy

It’s “Moving Day”….
Dave tries to push a
refrigerator over thick carpet.
He pushes hard, but it
doesn’t budge.
Steve pushes an identical
refrigerator across a wooden
floor. He also pushes hard,
and it moves easily.
Who does more “work”?
Steve
Steve moved the refrigerator….Dave didn’t.
Work – what is accomplished when a force acts on
an object and that force causes the object
to move over a certain distance
W = FdcosΘ
W = Work (Joule)
F = force (N)
d = distance (m)
Θ= angle between
the force and
distance
Is work being done by the
man on the groceries by
holding it in place?
Is work being done by the
Earth on the groceries?
The net work in the vertical
direction on the groceries is
zero (net force = 0)
Is work being done
by the man on the
groceries if he walks
forward?
No…his force on
the bag is
perpendicular to
motion
Work is a scalar quantity -- it only has
magnitude. However, there is a difference
between negative work and positive work.
The hammer does
positive work on the
nail (same direction
as displacement)
The nail does
negative work on
the hammer
(opposite direction
of displacement)
A person pulls a 50-kg crate 40 m along a horizontal
floor with a constant force of 100 N at an angle of 37 .
The floor is rough and exerts a friction force of 50 N.
Determine the work done by each force acting on the
crate and the net work being done on the crate.
If a net work is done on the bus above, what do we know
about the bus
•It moves
•A force is applied
•It accelerates
If an object’s speed changes, its kinetic energy changes
Kinetic energy – energy of motion (measured in joules)
KE = 1/2 mv2
KE = kinetic energy (J)
m = mass (kg)
v = speed (m/s)
Since work causes an object’s speed to change, it
causes its kinetic energy to change
(Work-Energy Theorem)
ΣW = ∆KE
ΣW = 1/2 mvf2 -1/2 mvi2
If ΣW (Wnet) > 0, KE increases (object speeds
up)
If ΣW (Wnet) < 0, KE decreases (object slows
down)
A baseball (m = 140 g) traveling 32 m/s moves a
fielder’s glove backward 25 cm when the ball is
caught. What was the average force exerted by
the ball on the glove?
Power
If you press a barbell over
your head with a constant
speed….
Are you doing work on
the barbell?
Yes
Is gravity doing work on
the barbell?
Yes (negative)
What is the net work on
the barbell?
Zero (no a, no ΣF)
If you faster reps, you feel
more fatigued. Why?
Your rate of doing work is
increasing.
Power = work / time
P=W/t
P = power (Watt)
W = work (J)
T = time (s)
1 Horsepower = 550 ft lb/s = 746 W
Since we know W = Fd,
P=Fd/t
P=Fv
P = power (W)
F = force (N)
v = average speed (m/s)
A pump is to lift 18.0 kg of water per minute to
a height of 3.60 m. Assuming the pump moves
the water at a constant speed, what power
rating (watts) should the pump motor have?
A car is traveling up a 10° incline at a constant
speed of 80 km/h. The car has a mass of 1400
kg and the frictional force of the road on the
car is 700 N. How much horesepower does
the motor need in order to move at that speed?
Forms of Energy
Energy
Mechanical
Kinetic
Nonmechanical
Potential
Gravitational
Elastic
Potential energy – energy associated with forces that depend
on the position of an object relative to its surroundings
Gravitational Potential Energy
An object gains energy when work is done to it to lift it off
the ground to a position above the ground.
W = Fadcosθ
If object is raised at a constant speed, Fa = Fg. Therefore,
W = Fd = PEG= mgh
PEg = gravitational potential energy (J)
m = mass (kg)
g = gravitational acceleration (m/s2)
h = height (m)
A roller coaster with a mass of 1000 kg travels up a 25
m hill and then back down 10 m. What is the potential
energy at the top of the hill and at the ending position
on the other side?
Elastic Potential Energy
A spring gains energy when work is done to it to
compress or stretch it from its natural position
Hooke’s Law
Fp = kx
Fs = -kx
k = spring constant (N/m)
x = amount of
compression/tension (m)
Fp increases as a spring is compressed, so average force
is F = ½(0 + kx) = ½ kx. Therefore,
W = Fd = PES = ½kx2
PEs = elastic potential energy (J)
k = spring constant (N/m)
x = amount of compression/tension (m)
A spring with a force constant of 5.2 N/m has a relaxed
length of 2.45 m. When a mass is attached to the end of
the spring and allowed to come to rest, the vertical length
of the spring is 3.57 m. Calculate the elastic potential
energy stored in the spring.
Principle of Conservation of Mechanical Energy
If only conservative forces are acting , the total
mechanical energy of a system neither
increases nor decreases in any process. It
stays constant – it is conserved.
Conservative forces – work done does not depend on
path; just the initial and final positions (gravitational
force, elastic force)
Nonconservative forces – work done depends on path
taken (friction force, air resistance, applied force,
tension force)
E = PE + KE
An object’s initial E at time 1 is equal to an
objects final E at time 2 assuming no
nonconservative forces acting.
E1 = E2
PE1 + KE1 = PE2 + KE2
If the original height of a rock is 3.0 m, calculate
the rock’s speed when it has fallen 2.0 m and the
speed right before it hits the ground.
A dart of mass 0.100 kg is pressed against the spring of a
toy dart gun. The spring (k = 250 N/m) is compressed 6.0
cm and released. If the dart detaches from the spring
when the spring reaches its natural length, what speed
does the dart acquire?
Law of Conservation of Energy
The total energy is neither increased nor
decreased in any process. Energy can be
transformed from one for to another, and
transferred from one object to another, but the
total amount remains constant.
In a process, mechanical energy may be “lost” or
“gained” due to the work that is done by
nonconservative forces, but it simply is converted into
or converted from another type of energy (heat, light,
electric energy, etc.)
WNC = ∆KE + ∆PE
The work done by nonconservative forces
(to produce heat, light, etc.) is equal to the
sum of the change in kinetic energy and the
change in potential energy.
You drop a 2.0-kg ball from a height of 2.0 m and it
bounces back to a height of 1.5 m. How much energy is
“lost” in the bounce? What is the ball’s speed at the
instant it leaves the ground after the bounce?
An SUV and a sports car are both moving 60 mph
down the highway.
Which one has more kinetic energy?
Which one has more momentum?
Our intuition tells us that the SUV has more
momentum because it is larger.
Is there any way to give the sports car more
momentum so that it eventually is the same as
the SUV?
How would we make their kinetic energies
equal?
Increase the speed of the sports car.
The same is true to make their momentums the
same – increase the velocity of the sports car.
Momentum - the product of an object’s mass
and velocity
p=mxv
p = momentum (kg x m/s)
m = mass (kg)
v = velocity (m/s)
Momentum is different from energy in that it
is a vector quantity. Direction matters!
How would we change the car’s momentum?
Accelerate the car by applying more force.
Σ F = ma
Σ F = m(∆v/∆t)
Σ F = (m∆v)/∆t
Σ F = ∆p/∆t
Newton’s Second Law – the net force applied
to an object is equal to the rate of the change
of momentum of an object
A tennis ball with a mass of 0.060 kg leaves a
tennis racket on the serve with a velocity of 55
m/s. If the racket is in contact with the ball for
0.004 seconds, what is the force that is exerted
on the ball?
Water leaves a hose at a rate of 1.5 kg/s with a
velocity of 20 m/s. If the water is aimed at a car
window which stops it, what is the force exerted
by the car on the water?
Impulse
When you participate in a water balloon toss,
how do you catch a water balloon?
“Soft Hands”
Your catch is spread over over a larger amount
of time.
If you were to abruptly catch the balloon, it
would break.
F = ∆p/∆t
F∆t = ∆p
To change an object’s momentum, if the time
decreases, the force has to increase (or vice
versa)
Impulse – the product of a force exerted on
an object and the time over which the force
acts
A 12-kg hammer strikes a nail at a velocity of 8.5 m/s and
comes to rest in a time interval of 8.0 ms. What is the
impulse given to the nail? What is the average force
acting on the nail?
Law of Conservation of Momentum
The total momentum of an isolated system of
objects remains constant.
In a system with numerous objects, the individual
momentums of the objects may change as the objects
interact with each other, but the sum of their momenta
remains constant.
mAvA0 + mBvB0 = mAvA +mBvB
Explosions
Initially, a system has no momentum. Then, some event
happens causing objects to move in various directions.
Regardless of the direction or motion, all of the resultant
momentums have to cancel out to equal zero (to match
the initial conditions).
Initial Momentum = Final Momentum
mv = mAvA + mBvB
What is the recoil velocity of a 5.0-kg rifle that
shoots a 0.020-kg bullet at a speed of 620 m/s?
Elastic Collisons
Initially, at least one object of a system has momentum. It
then hits another object causing each object to change its
velocity. All of the momentums of the objects after the
collision will equal the momentums of the objects before
the collision.
Initial Momentum = Final Momentum
mAvA0 + mBvB0 = mAvA +mBvB
Because the collision is elastic, not only is
momentum conserved, but all mechanical
energy is conserved.
½mAvA02 + ½mBvB02 = ½mAvA2 + ½mBvB2
mAvA02 + mBvB02 = mAvA2 + mBvB2
mAvA02 - mAvA2 = mBvB2 - mBvB02
mA(vA02 - vA2) = mB(vB2 - vB02)
mA(vA0 - vA)(vA0 + vA) = mB(vB - vB0)(vB + vB0)
mA(vA0 - vA)(vA0 + vA) = mB(vB - vB0)(vB + vB0)
Back to momentum…
mAvA0 + mBvB0 = mAvA +mBvB
mAvA0 - mAvA = mBvB - mBvB0
mA(vA0 - vA) = mB(vB - vB0)
Combining both equations…
vA0 + vA = vB + vB0
vA0 - vB0 = vB – vA
(for elastic collisions)
A ball of mass 0.440 kg moving east with a speed
of 2.20 m/s collides head-on with a 0.220-kg ball
at rest. If the collision is perfectly elastic, what
will be the velocity of each ball after the collision?
Inelastic Collisons
Inelastic collisions are ones where two separate
objects collide and stick together moving together
as one unit afterward. Momentum is conserved
in these types of collisions.
Initial Momentum = Final Momentum
mAvA + mBvB = (mA + mB)v
Mechanical energy is not conserved in these
types of collisions (deformation, heat, etc.)
A bullet is fired vertically into a 1.40-kg block of wood at
rest directly above it. If the bullet has a mass of 29.0 g and
a speed of 510 m/s, how high will the block rise after the
bullet becomes embedded in it?