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Part 3: Energy & Linear Momentum
3.1: Work
Energy & Work
If an object speeds up, energy is given to it. The amount of energy the object receives is the
work done on the object. This work is positive. The work is done by the application of force(s)
by other object(s).
If an object slows down, energy is removed from it. The amount of energy the object loses is the
work done on the object. This work is negative. The work is done by the application of force(s)
by other object(s).
Work: If a force is constant over a straight displacement, then
F
F
s

Units of Work & Energy:
mks

s
force
displacement of object
angle between force and displacement vectors
[Joule = J = N-m]
other common units
[1 ft.-lb. force = 1.3558 J]
[1 Btu = 1055 J]
[1 kW-hr = 3.6x106 J]
[1 cal = 4.184 J]
[1 Cal = 1kcal = 4.184 kJ]
[1 eV = 1.6x10-19 J]
3.2: Power
Power is the rate that work is done on an object. Thus, it is the rate that energy is supplied to or
removed from an object.
=
(average power)
(instantaneous power)
Units of Power:
mks
U.S.
[Watt = W = J/s]
[horsepower = hp]
[1 hp = 746 W]
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If a force is constant over a straight displacement, then
3.3 Kinetic Energy (Work-Energy Theorem)
Kinetic Energy
An object with mass m that moves at speed v has a kinetic energy of
K = (1/2) mv2
Work-Energy Theorem
Wnet = K = (1/2) mvf2 - (1/2) mvi2
Example:
Bob holds a 2-kg book stationary for 30 minutes. How much work has Bob done on the book?
Ans. zero
Example:
Bob raises a 2-kg book at a constant speed a distance of 1 meter with a vertical lifting force.
(a) How much work does Bob do?
(b) How much work does gravity do?
(c) What is the net work done on the book?
Ans. (a) 19.6 J (b) -19.6 J (c) zero
Example:
Bob applies a 20-N vertical lifting force to a 2-kg book initially at rest over a distance of 1 meter.
(a) How much work does Bob do?
(b) How much work does gravity do?
(c) What is the net work done on the book?
(d) What is the speed of the book at the end of the 1 meter?
(e) What is the average power delivered to the book?
Ans. (a) 20 J (b) -19.6 J (c) 0.4 J (d) 0.632 m/s (e) 0.127 W
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Example:
A hockey puck is sliding at 2 m/s on the ice. You skate up alongside the puck and apply a
constant 5-N force to the puck with your stick for 1 meter until you are hip checked into the
boards. Find the speed of the puck after this 1 meter if the coefficient of kinetic friction between
the puck and ice is 0.15. The mass of the hockey puck is 6 ounces (0.17 kg).
Ans. 7.74 m/s
3. 4 Potential Energy & Conservation of Energy
Potential Energy
A force is called conservative if the work it does on an object is independent of the path and only
depends on the initial and final positions. Such forces are gravity, the spring force, and the
electrical force. For a conservative force, there is a potential energy function (U ) that is a
function of position such that the work done by the force is W = Ui – Uf.
Gravity (near Earth’s surface)
U = mgh
h = height of object
Spring Force
U = (1/2)kx2
k = spring constant
x = distance spring is stretched/compressed from equilibrium position
Law of Conservation of Energy
Version 1.0: only gravity near Earth’s surface acts on object
(1/2)mvi2 + mghi = (1/2)mvf2 + mghf
Version 2.0: gravity near Earth’s surface & spring force
(1/2)mvi2 + mghi + (1/2)kxi2 = (1/2)mvf2 + mghf + (1/2)kxf2
Example:
A cart is at rest at the top of a curvy, frictionless track. The cart is 3 meters above the bottom of
the track. Find the speed of the cart at the bottom of the track if air drag is negligible.
Ans. 7.67 m/s
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Example:
Wiley Coyote has devised another trap for Roadrunner. Roadrunner is pecking at some seed at
the edge of a cliff. Wiley is 15 meters below near the base of the cliff, sitting atop a spring that
he has compressed 2 meters from equilibrium. The spring constant is 2000 N/m. Wiley
launches himself upwards so he can grab Roadrunner! (a) How fast is Wiley traveling half-way
up the cliff face? Wiley has a mass of 40 kg. (b) How high does Wiley ascend? Will he reach
Roadrunner? Neglect air drag in both calculations.
Ans. (a) 7.3 m/s (b) 10.2 m, nope
Example:
Energy Diagrams
(See worksheets handed out in class and posted on-line.)
Resistive Forces
Kinetic friction and air drag always do negative work on an object, thus the object loses kinetic
energy. This lost energy (Elost) is treated as a positive energy term on the final side of the
Conservation of Energy equation.
Kinetic Friction
Elost = fk s
s = distance traveled by object
Air Drag
Elost =  bv2 ds
[This is a difficult integration to do. In problems, Elost will be given or you will solve for it using
the Law of Conservation of Energy (see below). ]
Law of Conservation of Energy (Revisited)
Version 3.0: gravity near Earth’s surface & spring force & resistive forces
(1/2)mvi2 + mghi + (1/2)kxi2 = (1/2)mvf2 + mghf + (1/2)kxf2 + Elost
Version 4.0: gravity near Earth’s surface & spring force & resistive forces & any other forces
(1/2)mvi2 + mghi + (1/2)kxi2 + Wother = (1/2)mvf2 + mghf + (1/2)kxf2 + Elost
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Example:
Wiley Coyote is at it again. This time he is standing atop a flat cliff 40 meters above Roadrunner
who is pecking at some seed at the base of the cliff. Wiley pushes a 30-kg rock with a constant
horizontal force of 80 N a distance of 8 meters whereupon it leaves the edge of the cliff. The
coefficient of friction between the rock and the cliff top is 0.2. The rock loses 3000 J of energy
to air drag as it plummets towards the base. Find the impact speed of the rock. (Don’t worry,
the rock doesn’t hit Roadrunner as it travels a curved path, something that Wiley forgot to factor
into his calculations.)
Ans. 24.4 m/s = 55 mph
Gravity (in general)
U = -Gm1m2 / r
r = distance between objects’ gravitational centers
Law of Conservation of Energy (Revisited Again)
Version 5.0: general gravity
(1/2)mvi2 - Gm1m2 / ri = (1/2)mvf2 - Gm1m2 / rf
Example:
Find the launch speed required for an object if it is never to return to the Earth. This is the socalled escape speed for an object. Neglect drag from the atmosphere.
Ans. In the Conservation of Energy equation, set ri = RE, vf = 0, and rf = ∞. This gives
= 11.2 km/s = 25, 000 mph.
Example:
Pluto’s orbit around the Sun is very elliptical. At perihelion, Pluto is the closest it gets to the Sun
at a distance of 4.4368x109 km. Its speed there is 6.112 km/s. At aphelion, its farthest distance
from the Sun, Pluto is 7.3759x109 km away. How fast is Pluto moving at aphelion? The masses
of the Sun and Pluto are 1.9891x1030 kg and 1.305x1022 kg, respectively.
Ans. 3.678 km/s (60% of speed at perihelion)
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3. 5 Linear Momentum & Impulse
Momentum
mks units [kg-m/s = N-s]
More General Form of Newton’s Second Law
Suppose that a large force acts on an object during a small time interval. This force causes a
change in momentum, also called the impulse.
Impulse
In one-dimension, then
Impulse
The size of the average force is
Example:
A golf ball has a mass of 1.62 ounces (0.045 kg). When a typical amateur golfer hits the ball off
the tee with a driver, the head of the driver is in contact with the ball for 1/2000 second and the
ball leaves the tee at about 50 m/s.
(a) Find the impulse experienced by the ball.
(b) Find the average force exerted by the head of the driver on the ball.
Ans. (a) 2.295 N-s (b) 4590 N = 1030 lbs.
Example:
A pitcher throws a baseball at 90 mph (40.2 m/s). The batter hits the ball right back at the
pitcher at 110 mph (49.1 m/s). The ball is in contact with the bat for 0.7 ms. Find the impulse
experienced by the ball and the average force exerted by the bat on the ball. A baseball has a
mass of 5.125 ounces (0.145 kg).
Ans. impulse is 12.9485 N-s, average force is 18,500 N = 4150 lbs.
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3. 6 Conservation of Linear Momentum
Law of Conservation of Momentum
1-D Collisions:
m1v1i + m2v2i = m1v1f + m2v2f
(v’s can be + or -)
A collision is elastic if kinetic energy is also conserved.
A collision is inelastic if kinetic energy is not conserved.
A special case of an inelastic collision is a perfectly inelastic collision where the objects stick
together (v1f = v2f). This results in the most kinetic energy loss possible.
Example:
A 250-kg cannon is mounted on wheels. It fires a 1-kg ball with a muzzle velocity of 50 m/s.
(a) Find the recoil speed of the cannon.
(b) Does the ball gain or lose kinetic energy when the cannon is fired? How much?
(c) Does the cannon gain or lose kinetic energy when it is fired? How much?
(d) From where do the ball and cannon get their kinetic energy?
Ans. (a) 0.2 m/s (b) gains 1250 J (c) gains 5 J (d) gunpowder combustion
Example:
The cannon ball from the previous example immediately hits a stationary, 20-kg, foam target
also mounted on wheels. The ball becomes embedded in the target.
(a) Find the speed of the target after the collision.
(b) Does the ball gain or lose kinetic energy in the collision? How much?
(c) Does the target gain or lose kinetic energy in the collision? How much?
(d) Did all of the kinetic energy lost by the ball go into raising the kinetic energy of the
target? If not, what happened to this left over kinetic energy?
(e) How would you classify this collision, elastic, inelastic, or perfectly inelastic?
Ans. (a) 2.4 m/s (b) loses 1247 J (b) gains 58 J (d) no, 1189 J goes into sound, heat,
deformation of target (e) perfectly inelastic
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Rocket Propulsion
Rocket Equation
vex = exhaust speed of gas relative to rocket
vi = earlier speed of rocket when its mass is Mi
v = later speed of rocket when its mass is M
Average Thrust
= rate that fuel is burned [kg/s]
Example:
The Saturn V rocket was used in the Apollo missions to get us to the Moon. The first stage of
the rocket was used for take-off. It provided a thrust of 34.02 million Newtons and had a burn
time of 2.5 minutes. The gas was exhausted at approximately 2578 m/s from its nozzles.
(a) Find the mass of the fuel used in the first stage.
(b) The total mass of the rocket at launch was around 3 million kilograms. What percentage
of this mass was fuel for the first stage?
Ans. (a) 1,979,400 kg (b) about 65%
Example:
Suppose that you are living in the future on a space station and that you wish to travel to another
near-by station. You hop onto your rocket scooter and take off from rest. The initial mass of the
scooter is 200 kg. Its engine burns fuel at a rate of 1 kg per second and ejects the combusted gas
at 1000 m/s.
(a) If you want to reach a cruising speed of 500 m/s, how much long should you burn fuel?
How much fuel is burned?
(b) What is the average thrust provided by the scooter engine?
Ans. (a) 79 s, 79 kg of fuel (b) 1000 N