Honors Physics

By his power God raised the Lord from the dead, and he will raise us also. 1 Corinthians 6:14

 The scientific definition of Work: Using a force to move an object a certain distance ◦ The force and the direction of motion must be the same.  Work comes from the old English word, “Weorc”, which means activity.  Units: Joules (J) = N·m

 Work is important in science because it is related to energy.  A force that does not make an object move does no work.

 A product is multiplying 2 numbers.  A scalar is a quantity with NO DIRECTION.  Work is the Force times the displacement. ◦ The result is ENERGY, which has no direction. A dot product is a constraint on the formula, in which F and x MUST be parallel. To ensure that they are parallel we add the cosine.

A student with a mass of 80.0 kg runs up three flights of stairs in 12.0 sec. The student has gone a vertical distance of 8.0 m. Determine the amount of work done by the student to elevate her body to this height. Assume that her speed is constant. Why is it important that speed is constant? Does it matter how long it takes her to go up the stairs?

FORCE Displacement Force & motion are in the same direction.

FORCE Displacement Force opposes motion.

FORCE Displacement Force is perpendicular to motion.

 A teacher lecturing to her class?  A mouse pushing a piece of cheese across the floor with his nose? Not Work Work

The mouse is using a force to push the cheese a certain distance.

 Pushing a book across a desk?  Standing still holding a stack of books for an hour? Work Not Work

 Lifting a full backpack 1 meter from the ground?  Lifting an empty backpack one meter from the ground? Or…

 Lifting a full backpack 10 cm from the ground?  Lifting a full backpack one meter from the ground? Or…

 The work done on an object depends on the direction of the force applied and the direction of the motion.

 When the force and the motion are in the same direction, calculate work by multiplying the force and the distance.

 When the applied force and the motion of the object are NOT in the same direction, the applied force can be thought of as being two forces acting on the object at the same time.

 When the applied force and the motion of the object are NOT in the same direction, only the horizontal part of the applied force is used in the work equation. ◦ The vertical part of the applied force does no work on the suitcase.

This woman is applying a force at an angle theta. Only the Horizontal Component causes the box to move and thus imparts energy to the box. The vertical component (Fsin  ) does no work on the box because it is not parallel to the displacement.

If a man pushes a wheelbarrow 10 m with a force of 20 N, how much work has he done? Given: F = 20 N, d = 10 mFind: W

 The work done to lift an object equals the weight of the object multiplied by the distance it is lifted. ◦ Weight = mass x gravity

How much work does it take to lift an 8-kg backpack 1.5 meters? Given: m = 8 kg, d = 1.5 m Find: W

 We have learned Kinematics and Newton's Laws.  Let 's apply both to our new formula for work. 1.Start with Newton's Second Law 2. Use Kinematic #4 3. KINETIC ENERGY or the ENERGY OF MOTION

 Doing work on an object transfers energy to the object.  This helps scientists predict how an object will act when forces are applied to it.  Work done when you lift an object also increases the object’s energy.

 Work = The Scalar Product between Force and Displacement.  If you apply a force on an object and it covers a displacement you have supplied ENERGY or done WORK on that object.  Energy is the ability to do Work.

 If we impart work to an object it will undergo a change in speed and thus a change in Kinetic Energy.  Since both Work and Kinetic Energy are expressed in Joules, they are Equivalent Terms. " The Net Work done on an object is equal to the change in Kinetic Energy of the object."

Suppose a woman applies a 50 N force to a 25-kg box (initially at rest) at a 30  angle above the horizontal. She manages to pull the box 5 meters. a)Calculate the WORK done by the woman on the box b)The speed of the box after 5 meters if the box started from rest. a) Is this positive or negative work?

b) Suppose a woman applies a 50 N force to a 25-kg box (initially at rest) at a 30  angle above the horizontal. She manages to pull the box 5 meters. a)Calculate the WORK done by the woman on the box b)The speed of the box after 5 meters if the box started from rest.

A 58 kg skier is coasting down a 25° slope. A kinetic frictional force of 70 N opposes her motion. Near the top of the slope, the skier’s initial speed is 3.6 m/s. Ignoring air resistance, determine the final speed of the skier after a displacement of 57 m downhill. 569N

Given: m = 58 kgF f = 70 NV i = 3.6 m/s  x = 57 m

A 1000-kg car traveling with a speed of 25 mls skids to a stop. The car experiences an 8000 N force of friction. Determine the stopping distance of the car. Given: m = 1000 kgF f = NV i = 25 m/s V f = 0 m/s

Suppose you lift a mass upward at a constant speed,  v = 0 &  K=0. What does the work equal now? Since you are lifting at a constant speed, your Applied Force equals the Weight of the object you are lifting. Since you are lifting you are raising the object a certain “y” displacement or height above the ground. When you lift an object above the ground it has Potential Energy.

mg h Since this man is lifting the package upward at a constant speed, the kinetic energy is not changing. The work that he does goes into the Energy of Position or Potential Energy. All potential energy is considering to be energy that is stored.

The man shown lifts a 10 kg package 2 meters above the ground. What is the potential energy given to the package by the man? 196 J h