Part 1: Introduction & 1-D Kinematics



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

[pic]

F force

s displacement of object

( angle between force and displacement vectors

Units of Work & Energy: mks [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.

[pic]=[pic] (average power)

[pic] (instantaneous power)

Units of Power: mks [Watt = W = J/s]

U.S. [horsepower = hp] [1 hp = 746 W]

If a force is constant over a straight displacement, then

[pic]

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

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

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

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 so-called 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

[pic] = 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)

3. 5 Linear Momentum & Impulse

Momentum [pic] mks units [kg-m/s = N-s]

More General Form of Newton’s Second Law [pic]

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 [pic]

In one-dimension, then

Impulse [pic]

The size of the average force is

[pic]

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.

3. 6 Conservation of Linear Momentum

Law of Conservation of Momentum

[pic]

1-D Collisions: m1v1i + m2v2i = m1v1f + m2v2f (v’s can be + or -)

A collision is elastic if kinetic energy is also conserved.

[pic]

A collision is inelastic if kinetic energy is not conserved.

[pic]

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

Rocket Propulsion

Rocket Equation [pic]

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 [pic]

[pic] = 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

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