reporters khlaire abegail m ruperez willowy coleen d boncales hanz apple m coscos julius sulapas n.
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WORK, POWER, ENERGY PowerPoint Presentation
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  1. WORK, POWER, ENERGY Reporters: KhlaireAbegail m. RuperezWillowy coleen d. boncaleshanz apple m. coscosjuliussulapas

  2. Work -quantity that exist whenever a force acting upon an object causes a displacement. W=F(cos ) d (Eqn 8.1) 3 key words *force *displacement *cause. Unit of work “J” standard metric unit

  3. 1 joule = 1 newton * 1 meter 1 J = 1N * m ~Non-standard units of work~ • ft-pound • Kg*m/s²*m • Kg*m²/s² * work is a force acting upon an object to case a displacement. When a force acts to cause an object to be displaced, 3 quantities( force, displacement and angle) must be known in order to calculate the amount of work. Maximum Work -work is maximum if the force applied is completely along the direction of the displacement.

  4. A 100 N force is applied to move a 15 kg object a horizontal distance of 5 meters at constant speed. Example of Work (without angle) ANSWER W = (100 N) * (5 m)* cos(0 degrees) = 500 J The force and the displacement are given in the problem statement. It is said (or shown or implied) that the force and the displacement are both rightward. Since F and d are in the same direction,the angle is 0 degrees.

  5. Example of Work ( with angle) A 100 N force is applied at an angle of 30° to the horizontal to move a 15 kg object at a constant speed for a horizontal distance of 5 meter. ANSWER W = (100 N) * (5 m) * cos(30 degrees) = 433 J The force and the displacement are given in theproblem statement. It is said that the displacement is rightward. It is shown that the force is 30 degrees above the horizontal. Thus, the angle between F and d is 30 degrees.

  6. Example Lifting Quarter Pounders If I lift a quarter-pound hamburger with a force of 1 newton up a distance of one meter, how much work did I do? Answer Given: Force applied F 1 Newton Distance traveled d 1 meter Find: Work done W Solution Substituting the given values directly to W=F x d. Work = F x d = 1 newton x 1 meter =1 newton – meter = 1 joule Answer: I did one joule of work.

  7. Work done against gravity W = mg (will be our f) W = F x d = mg x h Example 8.2 “Sagingpara climbing” Eating banana enables a person to perform about 4.0x10^4 J of work. How high can a 60-kg woman climb if energized by eating a banana? Answer Given: work that can be done from eating banana W 4.0 x 10^4 Mass of woman m 60 kg Find: How high a woman can climb from eating a banana (h)?

  8. since the motion is along the vertical, we can use the equation for work done against gravity, then solve for h. From W = mgh Simple Machines ---A machine is anything that makes easier. 2 ways machines help man: *Some machine help man apply a “greater effective force” on the object despite applying a lesser amount of force . *Some machine help move an object over longer distance even if your work is done over shorter distance.

  9. (Mechanical Advantage) MA = force on the load / force into machine = Fout / Fin Efficiency measures how much of the work you put in gets to the load. Efficiency = useful work output/ total work input =Wout/ Win *3 simple machine of lever groups -levers -wheels -axles and pulleys

  10. Lever- is a board balanced on a point of support that can be rotated to lift something. *3 basic types of levers • First-class (the longer the better) • Second-class (short but powerful) • Third class (distance matters) *First-class lever, the fulcrum is between the load and the applied force. *second-class lever, the load lies between the fulcrum and the applied force. *third-class lever, the effort is applied between the load and the fulcrum, your effort will be greater, but you can move objets over longer distances.

  11. The Wheel and Axle Wheel and Axle essentially a modified lever, but using two circles, one smaller (axle) and one bigger (the wheel) attached in the middle. It makes it easier to push a car forward than if you had to slide it along the ground with no wheels. The axle must move with the wheels. Every small turn of the axle corresponds to a bigger turn from the wheel. Pulley is a rope over a cylinder, which changes the direction of the force, you pull the rope down and the object moves up.

  12. 3 Machines make up the inclined plane group. -ramps -wedges -screws Inclined Plane makes it easier to move a heavy object up but lessening the force but making you push over a longer distance than you would need to lift it. Inclined Plane is also known as Ramps. Wedges has 2 parts to consider : b (base) and h (height)

  13. The Screw has 2 important parts to consider: p (pitch) and r (radius of the circular head) Example “Gaining Leverage” Using a lever how much force is needed to pick up a 100-newton boulder? Given: weight of the boulder Fi100N Big distance di1m Small distance do 0.01 Find: force needed to pick up the boulder Fi.

  14. Solution little force x big distance = big force x little distance Fi x di = Fo x do Fi x 1m = 100N x 0.01 m Fi = 1N Example 8.4 “Pulley, pulley” If you put in 200 joules of work to lift a 100 N box 1 meter with a pulley system. What is the machine’s efficiency? Given: Total Work Input Win 200J Resisting force Fout 100N Distance moved by the applied force dout 1m

  15. Find: Efficiency Solution Efficiency = Wout / Win *solve for Wout Wout = Fout x dout = 100N x 1m = 100J *solve for efficiency 100N x 1 meter/ (200 N m) = 0.5 *Percentage Efficiency = 0.5 * 100% = 50%

  16. Power – amount of work done per unit time. *unit of power 1 watt = 1 J/s Horsepower (hp) is the traditional unit of power in engineering. 1 hp = 746 W =0.746 kW 1 kW = 1.34 hp

  17. Example 8.5 An electric motor with an output of 15 kW provides power for the elevator of a six-story building. If the total mass of loaded elevator is 1000 kg, what is the minimum time needed for it to rise 30 m from the ground floor to the top floor? Given Power output 15 kW (P) Total mass of elevator 1000 kg (m) Height raised 30 m (h) Find: Time to rise (t)

  18. 15 kW * 1000 watts / 1 kw = 15, 000 watts *solve for t= W/P w = mgh (work done) Substitute t=W/P=mgh/P t= 1000 kg * 9.8 m/s² * 30m 15,000 Watts *since 1 watt = 1N*m = 1 kg-m/s²-m = 1 kg-m²/s² t=1000 kg * 9.8 m/s² * 30m 15,000 kg – m²/s = 20s The time needed for the 15 kW motor to raise a 1000-kg elevator by 30 m is 20s.

  19. Energy- the ability to do work. 2 kind of energy *potential energy -is energy that is stored and waiting to be used. *kinetic energy - is energy in movement. Has 3 types the Vibrational (ex. When strumming the guitar), Rotational (ex. Biking – spinning wheels) and Translational (ex. Surfing- the surfer’s movement) *Formula KE=0.5 mv²

  20. Example: If you throw a cheeseburger deluxe weighing 0.1 kg at 10 meters per second, how much kinetic energy have I given it? Given Mass of the burger m 0.1 kg Velocity of the burger v 10 m/s Find: Kinetic Energy given to the burger (KE) Solution KE = 0.5 (0.1 kg) 8 (10 m/s)² = 5 kg-m²/s²= 5 N-m = 5 J Answer: You gave that cheeseburger 5 joules of energy by throwing it.

  21. 2 forms of potential energy *Gravitational Potential Energy -the energy stored in an object as the result of its vertical position. PE grav = mass*g*height (eqn. 8.12) = w * h (eqn. 8.13) Example: (formula: eqn. 8.13) If you lift a 1-Newton cheeseburger deluxe 1 meter how much gravitational potential energy have you give it?

  22. Solution PEgrav = weight (force to lift) x height = 1 newton x 1 meter = 1 joule Answer: I gave that burger 1 joule of gravitational potential energy. Example: (formula: eqn. 8.12) What is the gravitational potential energy of a 0.5-kg ball soon to be dropped from a 15 m high building? Given Mass of the ball m 0.5 kg Height of the ball h 15 m Find: Gravitational potential energy of the ball (PEgrav)

  23. Solution Pegrav = mgh = 0.5 kg (9.8m/s²) (15m) = 73.5 J Answer:73.5 J *Elastic Potential Energy -the energy stored in elastic materials as the result of their stretching or compressing. -can be stored in rubber hands, bungee chords and trampolines.

  24. Question 1: A compressed spring has potential energy of 20 J. If the spring constant of the spring is 200 N/m, find the displacement of the spring? Solution: Given: Potential energy P.E = 20 J,      Spring Constant k = 200 N/m, The Potential energy is given by P.E = 1/2 kx² X= X= X=0.4 m