5.4 The Work-Energy Theorem and Kinetic Energy ~Net work done on an object is equal to the change in the kinetic energy of the object When a net external force does work on and object, the kinetic energy of the object changes according to

5.4 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the .056 N force acts on the probe through a displacement of 2.42×109m, what is its final speed?

5.4 The Work-Energy Theorem and Kinetic Energy

5.4 The Work-Energy Theorem and Kinetic Energy

THE WORK-ENERGY THEOREM 5.4 Work, Energy, and Power THE WORK-ENERGY THEOREM Work done by one object on another Net force on an object and relates the net work done on an object to the change in kinetic energy

THE WORK-ENERGY THEOREM 5.4 Work, Energy, and Power THE WORK-ENERGY THEOREM

5.4 Work, Energy, and Power Question #27 A 2.0 X 103 kg car moves down a level highway under the actions of two forces. One is a 1140 N forward force exerted on the wheels by the road. The other is a 950 N resistive force exerted on the car by the air. Use the work-kinetic energy theorem to find the speed of the car after it has moved a distance of 20.0 m assuming the car stars from rest. 1.9 m/s

5.4 Work, Energy, and Power Question #28 A 2.10 X 103 kg car starts from rest at the top of a driveway 5.0m long that is sloped at 20.0° with the horizontal. If an average friction force of 4.0 X 103 impedes the motion, what is the speed of the car at the bottom of the driveway 3.8 m/s

5.4 Work, Energy, and Power Question #29 A 10.0kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. the pulling force is 100 N parallel to the incline, which makes an angle of 15.0 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following: The work done by the Earth’s gravity on the crate. -1.9 X 102 J

5.4 Work, Energy, and Power Question #30 A 10.0kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. the pulling force is 100 N parallel to the incline, which makes an angle of 15.0 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following: The work done by the force of friction on the crate. -2.8 X 102 J

5.4 Work, Energy, and Power Question #31 A 10.0kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. the pulling force is 100 N parallel to the incline, which makes an angle of 15.0 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following: The work done by the puller on the crate. 7.5X 102 J

5.4 Work, Energy, and Power Question #32 A 10.0kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. the pulling force is 100 N parallel to the incline, which makes an angle of 15.0 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following: The change in kinetic energy of the crate. 2.8 X 102 J

5.4 Work, Energy, and Power Question #33 A 10.0kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. the pulling force is 100 N parallel to the incline, which makes an angle of 15.0 ° with the horizontal. Assuming the coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 7.5m, find the following: The speed of the crate after it is pulled 7.5m. 7.6 m/s

Power The rate of energy transfer Energy used or work done per second Remind students that W = Fd, and ask them to substitute this for W in the power formula. Then ask what d/t represents. At this point, move on to the next slide, which shows the alternative form of the power equation (P = Fv).

DEFINITION OF AVERAGE POWER Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work.

Power SI units for power are J/s. Called watts (W) Equivalent to kg•m2/s3 Horsepower (hp) is a unit used in the Avoirdupois system. 1.00 hp = 746 W Be sure students understand that P = Fv is not a new definition. It is simply a different but equivalent formula that makes calculations easier in some cases. Do not just show the units. Ask students to figure them out. It should be easy for them to get J/s but the basic units of kg•m2/s3 will be more difficult. This provides a good opportunity to review the units for joules and newtons. The horsepower was based on the work a good horse could do lifting coal out of a mine. A good horse could lift 275 pounds of coal at 2.0 ft/s, so it could do 550 ft•lb/s. This is equivalent to 746 J/s or 746 W.

5.4Power Power = force X speed Power = watt = J/s

5.4 Power

5.4 Power Question #34 A 1.0 X 103 kg elevator carries a maximum load of 800.0 kg. a constant frictional force of 4.0 X 103 N slows the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of 3.00 m/s. 66 kW

5.4 Power Question #35 A 1.50 X 103 kg car accelerates uniformly from rest to 10.0 m/s in 3.00s. What is the work done on the car in this time interval? What is the power delivered by the engine in this time interval? 7.50 X 104 J 2.50 X 104 W

5.4 Power Question #36 A car with a mass of 1.50 X 103 kg starts from rest and accelerates to 18.0 m/s in 12.0 s. Assume that air resistance remains constant at 400.0 N during this time. What is the average power developed by the engine? 2.38 X 104 W

5.4 Power Question #37 A rain cloud contains 2.66 X 107 kg of water vapor. How long would it take for a 2.00 kW pump to raise the same amount of water to the cloud’s altitude, 2.00km? 2.61 X 108 s