Tuesday, 18 April 2006

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Pengalaman Belajar di SMAN BI 1 Banjar

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Newton's Third Law of Motion


SUBJECT: Physics
TOPIC: Force and Motion
DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion.
CONTRIBUTED BY: Carol Hodanbosi
EDITED BY: Jonathan G. Fairman - August 1996

The third law of motion states that if a body exerts a force on a second body, the second body exerts a force that is equal in magnitude and opposite in direction to the first force. So for every action force there is always a reaction force. No force can occur by itself
The book lying on the table is exerting a downward force on the table, while the table is exerting an upward reaction force on the book. Because the forces are equal and opposite, the book remains at rest. Notice also that the table legs are in contact with the floor and exert a force downward on it, while the floor in turn exerts an equal and opposite force upward.
Figure showing a book on a table and the force pairs on the

 book and the legs.
Questions for you to consider:

  • If forces are always equal and opposite in action and reaction, how is it possible for an object to accelerate?
    (Answer)

  • Explain, in detail, using the third law of motion, how a person is able to walk forward.
    (Answer)

  • There is a classic problem that physicists like to ask students. A horse is pulling a carriage on a level ground. The horse knows the third law of motion. He tells the carriage that he will exert a force forward, and the carriage will exert a force equal to the horse's force but in opposite directions. Therefore, the horse explained, he can never pull the carriage forward. Can you explain to the horse that he is mistaken? How is he able to pull the carriage forward?
    (Answer)



  • Sumber:

    NASA

    Saturday, 18 March 2006

    Fisika SMA

    Pengalaman Belajar Fisika di SMAN BI 1 Banjar

    Ada Apa Dengan Fisika?

    The First and Second Laws of Motion


    SUBJECT: Physics
    TOPIC: Force and Motion
    DESCRIPTION: A set of mathematics problems dealing with Newton's Laws of Motion.
    CONTRIBUTED BY: Carol Hodanbosi
    EDITED BY: Jonathan G. Fairman - August 1996
    Newton's First Law of Motion states that a body at rest will remain at rest unless an outside force acts on it, and a body in motion at a constant velocity will remain in motion in a straight line unless acted upon by an outside force.
    If a body experiences an acceleration ( or deceleration) or a change in direction of motion, it must have an outside force acting on it. Outside forces are sometimes called net forces or unbalanced forces.
    The property that a body has that resists motion if at rest, or resists speeding or slowing up, if in motion, is called inertia. Inertia is proportional to a body's mass, or the amount of matter that a body has. The more mass a body has, the more inertia it has.
    The Second Law of Motion states that if an unbalanced force acts on a body, that body will experience acceleration ( or deceleration), that is, a change of speed. One can say that a body at rest is considered to have zero speed, ( a constant speed). So any force that causes a body to move is an unbalanced force. Also, any force, such as friction, or gravity, that causes a body to slow down or speed up, is an unbalanced force. This law can be shown by the following formula
    F= ma
    • F is the unbalanced force
    • m is the object's mass
    • a is the acceleration that the force causes
    If the units of force are in newtons, then the units of mass are kilograms and the units of acceleration are m/s2. If the units of force are in pounds (English), then the units of mass are in slugs, and the units of acceleration are ft/s2.
    Motion of an object that is not accelerated (moving at a constant speed and in a straight line) can be found using the formula
    d= v t

    • d is the distance traveled
    • v is the rate of motion (velocity)
    • t is the time
    Some sample problems that illustrates the first and second laws of motion are shown below:
    Example 1
    If the speed of sound on a particular day is 343 m/s, and an echo takes 2.5 seconds to return from a cliff far away, can you determine how far the cliff is from the person making the sound?
    An echo is a sound that travels out and back. It take 2.5 seconds for this trip, which is twice the distance to the cliff. Therefore, it only takes 1.25 seconds for the sound to reach the cliff. By substitution,
    d = v t
    d = (343 m/s) (1.25 s)
    d = 429 m

    Example 2
    If an unbalanced force of 600 newtons acts on a body to accelerate it at +15 m/s2, what is the mass of the body?
    F = ma
    m=F/a
    m = 600n/15 m/s2
    m= 40 kg
    Exercises:
    1. If a car is traveling at 50 km/hr along a straight line, how many meters does it travel in 10 seconds?
      (Answer)
    2. A force of 5000 newtons is applied to a 1200 kg car at rest. What is its acceleration?
      (Answer)
    3. A 10 kg body has an acceleration of 2 m/s2. Find the net force acting on the body.
      (Answer)
    4. An empty truck with a mass of 2500 kg has an engine that will accelerate at a rate of 1.5 m/s2. What will be the acceleration when the truck has an additional load of 1500 kg ?
      (Answer)
    5. A box resting on a table has a mass of 5.0 kg.
      1. What is its weight?
      2. What will be its acceleration when an unbalanced horizontal force of 40 newtons acts on it?
        (Answer)

    6. What is the weight of an object that has a mass of 60 slugs?
      (Answer)
    7. A net force of 75 pounds acts on a body of 25 slugs. The body is initially at rest. What is its acceleration during the action of the force?
      (Answer)
    8. What is the mass of a 185 pound man? If a 100 pound horizontal net force acts on the man while he is sitting on a wooden floor, what will his acceleration be?
      (Answer)

    Saturday, 18 February 2006

    Fisika SMA

    Pengalaman Belajar Fisika di SMAN BI 1 Banjar

    Ada Apa Dengan Fisika?

    Lesson Plan: Graphing Data from a Spreadsheet


    SUBJECT: Technology
    TOPIC: Presenting Data
    DESCRIPTION: A set of problems dealing with graphing data.
    CONTRIBUTED BY: Carol Hodanbosi
    EDITED BY: Jonathan G. Fairman - August 1996

    Purpose:
    To graph real data to input into a computer spreadsheet
    Objective:
    To illustrate a distance versus time graph using student data from running various distances.
    Materials:
    • computer spreadsheet with data from previous lab activity
    • computer graphing utility
    Procedure:
    1. Each student will be graphing the average of their classes running speed data, putting distance (from 0 meters to 20 meters on the vertical or Y axis, and time on the X axis ( in seconds), in 0.1 second intervals.
    2. Plot the values for your class, using the average time for the 5, 10, and 15 meter distances.
    3. Plot a fourth point using a zero distance, and estimating what the time would be for that distance.
    4. Draw a straight, best-fit line through the data points.
    5. Using the points for the 15 meter time and the zero point as endpoints, calculate the slope of the line segment.
    6. What does the slope represent?
    7. Can you predict the slope of the line if the distance were extended to a much greater value, such as 10,000 meters (10k), or 40,000 meters?

    A blank graph of distance versus time
    Sumber:
    NASA