Presentation on theme: "In this chapter you will: Recognize that work and power describe how the external world changes the energy of a system. Relate force to work and explain."— Presentation transcript:
In this chapter you will: Recognize that work and power describe how the external world changes the energy of a system. Relate force to work and explain how machines ease the load.
Chapter 10 Sections Section 10.1: Energy and Work Section 10.2: Machines
Section 10.1 Energy and Work Objectives Describe the relationship between work and energy. Calculate work. Calculate the power used.
WORK AND ENERGY Conserved Properties – properties that are the same before and after an interaction. Examples are energy and momentum. Work – is the product of the force and the object’s displacement. It is equal to the constant force exerted on an object in the direction of motion times the object’s displacement. It is the transfer of energy by Mechanical means. It is denoted by W. It is measured in Joules. W = Fd Energy – the ability of an object to produce a change in itself or in its surroundings.
WORK AND ENERGY Kinetic Energy – is equal to ½ times the mass of an object times the speed of the object squared. It is denoted by KE. It is measured in Joules. KE = ½ mv 2 Work Energy Theorem – states that work is equal to the change in Kinetic energy. W = ΔKE James Prescott Joule – physicist that established the relationship between work done and the change in energy that results. The unit of Energy is named after him. Joule – unit of Work. One Joule equals 1 Newton meter which equals 1 kg*m 2 /s 2 (1 J = 1 Nm = 1 kg*m 2 /s 2 ).
WORK AND ENERGY If the external world does work on a system then “W” is positive and the energy of the system increases. If the system does work on the external world then “W” is negative and the energy of the system decreases.
CALCULATING WORK Work is done on an object only if it moves. If you hold an object in place you do not do any work. Also if you move an object at constant velocity at a constant height you do no work (because the force is up and the motion is sideways not in the same direction of the force). ***Work is only done when the force and displacement are in the same direction.*** A force displacement graph can give you a picture of the work done. The area under the curve of a force displacement graph represents the work done. ***If Force and Displacement are at right angles then W = 0.
CALCULATING WORK Remember that you can replace a force by it’s component parts (the x and y parts of the Vector Force). The work you do when you exert a force at an angle to a motion is equal to the component of force in the direction of the motion times the distance moved. The magnitude of the component of force F acting in the direction of motion is found by multiplying the force F by the cosine of the angle between F and the direction of motion.
CALCULATING WORK Work at an Angle – is equal to the product of the force and displacement times the cosine of the angle between the force and the direction of the displacement. W = Fd cos The work done by Friction acts in the direction opposite that of motion. So the work done by Friction is Negative. Negative work reduces the Kinetic Energy of the system.
CALCULATING WORK Do Example Problem 1 p. 261 W = FdW = ΔKE W = 4.5(.15).675 J = ΔKE W =.675 Joules Do Practice Problems p. 261 # 1-3 Do Example Problem 2 p. 262 W = Fd cos W = 255(30) cos 25° W = 7650(.90631) W = 6,933.25 J Do Practice Problems p. 262 # 4-8
CALCULATING WORK Work is the area under the curve in a graph of Force versus Displacement. If several forces are exerted on a system, calculate the work done by each force and then add the results.
POWER Power – is the work done divided by the time taken to do the work. It is denoted by “P”. It is measured in Watts. P = W / t ; P = Fd / t ; P = Fd cos θ / t ; P = Fv Watt – the unit of Power. It is equal to 1 Joule per second (J/s). Power is often measured in Kilowatts since a watt is so small. 1 KW = 1000 watts. Do Example Problem 3 p. 264 P = W / t = Fd / t P = 12,000(9) / 15 P = 108,000 / 15 P = 7,200 Watts or 7.2 KW also 7,200 J/s Do Practice Problems p. 264 # 9-14
POWER Since Displacement divided by time (d / t) = velocity we can find Power using P = Fv Do 10.1 Section Review p. 265 # 15-23