P11.-WORK.ppt
- 1. WORK MR. CYRILL N. CADANO
- 2. should understand the definition of work, including when it is positive, negative, or zero, Calculate the work done by a specified constant force on an object that undergoes a specified displacement. Describe relationship between work, energy, and force Students will explore the concept of work by taking notes and solving practice LEARNING OBJECTIVES
- 3. • Work is done when a force is applied to move an object. As far as Physics is concerned… When a force is applied to an object, over a distance, then work is done on that object.
- 4. Work-Energy Relationship • The net work done in a body is equivalent to Kinetic energy (Work-Kinetic Energy Theorem). • This allows us to think of kinetic energy as the work an object can do while it changes speed.
- 5. Is Work Being Done? • A boy pushes a lawn mower. – • Yes The mower is being pushed by the boy. Energy is transferred in the process. The displacement of the object and the force are in the same direction.
- 6. Is work being done? A waiter serving his customers. Yes and No. As he lifts the dishes he is doing work on them. As he holds them –no work is done Even if he is moving the dishes horizontally at constant speed, no work is done. More about that later…AND when he places them back down, negative work is done. More about that later…
- 7. Work is a scalar quantity and has positive or negative displacement. • Work is positive when the component of the force is the same direction as the displacement. • For example, when you lift a box, the work done by the force you exert on the box is positive because the force is upward, in the same direction as the displacement.
- 8. • Work is negative when the force is in the direction opposite the displacement. • For example, the force of kinetic friction between a sliding box and the floor is opposite the displacement of the box, so the work done by the force of friction on the box is negative.
- 9. • In order for positive work to occur, the force and displacement must be in the same direction.
- 10. • Assume you forgot to set the parking break and your car starts rolling down a hill. • You try in vain to stop it by pulling as hard as you can on the bumper, but the car keeps on moving forward. • You exert a force on the car opposite to the direction of travel. The distance traveled in the direction of the force is negative, you do negative work on the car. • But the car is pulling you in the direction of travel with a force of equal magnitude (Newton's third law). The car is doing positive work on you.
- 11. Physics Text (pg 156) - Serway/ Faughn
- 12. For each of the following, indicate whether the work done on the second object will have a positive or negative value. • The road exerts a friction force on a speeding car skidding to a stop. • Work has a negative value because the work caused the moving car to slow down.
- 13. Work Sign tips: • Force is in direction of motion + work • Force opposes motion - work • Force is 90o to motion no work • Object not in motion no work
- 14. Is Work done while carrying an object at constant speed? • NO – The force acting on the object must be in the same, or opposite direction to the object’s displacement or at some angle less that 90o (perpendicular to the object).
- 16. In Summary… • When a force moves a body on which it acts in the direction of the force, we say it has done work.
- 17. Calculating Work • The work done by a force is found by multiplying the force by the distance it has moved the body in the direction of the force. • Equations: W = Fd F = W/d d = W/F Units of Work: Joules 1J = 1 Newton-Meters Also Foot-Pound Joule = 0.737562149 foot pounds
- 18. • A student’s backpack weighs 30N. She lifts it from the floor to a shelf 1.5m high. How much work is done on the pack full of books. • Since the backpack weighs 30N, the force needed to lift it off the ground is equivalent to the 30N it weighs. (Lifting it requires a force equivalent to the force of gravity attracting it to the Earth.) F = 30N d = 1.5m W = F x d = (30N)(1.5m) = 45N.m = 45 J
- 19. A person pulls a block 2 m along a horizontal surface by a constant force F = 20 N. Determine the work done by force F acting on the block. Force (F) = 20 N Displacement (s) = 2 m Solution : W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule
- 20. A body falls freely from rest, from a height of 2 m. If acceleration due to gravity is 10 m/s2, determine the work done by the force of gravity! Known : Object’s mass (m) = 1 kg Height (h) = 2 m Acceleration due to gravity (g) = 10 m/s2 Solution : W = F d = w h = m g h W = (1)(10)(2) = 20 Joule
- 21. A force F = 10 N accelerates a box over a displacement 2 m. The floor is rough and exerts a friction force Fk = 2 N. Determine the net work done on the box. Known : Force (F) = 10 N Force of kinetic friction (Fk) = 2 N Displacement (d) = 2 m Solution : Work done by force F : W1 = F d cos 0 = (10)(2)(1) = 20 Joule Work done by force of kinetic friction (Fk) : W2 = Fk d = (2)(2)(cos 180) = (2)(2)(-1) = -4 Joule Net work : Wnet = W1 – W2 Wnet = 20 – 4 Wnet = 16 Joule
- 22. Activity 2.3 1. How much work is done when a 50N crate is pushed along a floor a distance of 10m? 2. A carpenter lifts a 45 kg beam 1.2m. How much work is done on the beam? 3. Work done by Tom and Jerry so the car can move as far as 4 meters. Forces exerted by Tom and Jerry are 50 N and 70 N.