1 Work is a Scalar quantity – does not have a direction.Chapter 5 Work and Energy 5.1 Work •Term ‘work’ has special meaning in science – work is done ONLY if a force moves an object. •The distance an object moves ALSO must be in the same direction as the force applied. Equation: Work = Force x Distance Work = F x d (N) x (m) SI Units of work - Newton – meter called a ‘Joule’ Units of Work = Joules (J) Work is a Scalar quantity – does not have a direction.

2 A force of 22N is applied over a distance of 5.0m W = F • d A force of 22N is applied over a distance of 5.0m W = F • d = 22N • 5.0m = 110 J 2. A force of 250,000N is applied over a distance of 15 km. = 250,000N • 15,000m = x 109 J

3 Is Work Being Done? It is only the horizontal component of the tension force in the chain that causes Fido to be displaced to the right.

4 Requires the same amount of work.The amount of work done by a force on any object is given by the equation Work = F * d * cos (θ)

5 For work to be done, the Force must be in the direction of the motion.If it is at an angle, the work done is: Work = Force • (cos θ) • displacement W = F • (cos θ) • d How much work do you do if you push a lawn mower 25m with a force of 85N at an angle of 30° ?   W = F • (cosθ) • d = = J 2. How much force would it take to do the same amount of work if the angle was 60° ? 1840 J = F x cos ϴ x d F = J = 147 N cos ϴ x d

6 5-2 Kinetic Energy Energy comes in many forms.Electrical, chemical, heat, and atomic are just a few examples. A good definition of Energy is the ability to do work. Work is a form of energy – energy is a form of work. The Kinetic Energy (KE) of an object is the amount of “work” stored by that object due to its motion. Formula: KE = ½ mv2 Units: J (Joules) The velocity of the object is squared and therefore the most influential component.

7 Example #1 A bowling ball ( m = 6.0 kg ) is traveling at 4.0 m/s. Find its K.E. at this speed. K. E. = ½ mv2 = ½ • 6.0 kg • (4.0) 2 = 48 J

8 More Practice - 1. What is the K. E. of a chicken ( m = 1.5 kg ) that is caught in a tornado (v = 130 m/s)? KE = ½ m x v2 = J or 13,000 J (2 sig figs)  2. What is the K.E. of a rain drop ( m = 6.5x10-4 kg ) as it hits your face? ( vy = 12 m/s ) = J or 4.7 x J 3. What is the K.E. of an excited Irish Setter (m = 34 kg ) that leaps on you with a velocity of 6.0 m/s ? = 612 J

9 5-2 continuation: WORK – ENERGY THEOREMWhen work is done on an object, the Net Work is converted to K.E. Wnet = ∆ K. E. Ex. ( p.176 ) 1. A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 45 N. How far must the student be pushed, starting from rest, so that her final K.E. is 352 J? Given F = and KE = 352 J Find ∆X (d) = ? Formula: Wnet = ∆ KE F ● ∆X = ∆ KE Wnet = F ● ∆X ∆X = ∆ KE / F = 7.8 m   

10 5-2 continuation: WORK – ENERGY THEOREMWhen work is done on an object, the Net Work is converted to K.E. Wnet = ∆ K. E. Ex. ( p.176 )  2. A 2.0 x 103 kg car accelerates from rest under the action of two forces. One is a forward force of 1140 N provided by traction between the tires and the road. The other is a 950 N resistive force due to various frictional forces. Use the WORK – ENERGY theorem to determine how far the car must travel for its speed to reach 2.0 m/s.  Given: m= 2,000 kg F = 1140 Ff = 950N V = 2.0 m/s H.W. – GR 5.2 and do problems p. 176 # 4 & 5

11 How far will it Skid?

12 Example: Fill out the following table for a car ( m = 1,600 kg ) Since K.E. depends on the square of the velocity, it increases dramatically as velocity increases. Example: Fill out the following table for a car ( m = 1,600 kg ) at different velocities; μs = 0.9 and Fs = μs x FN and Wbreaks = Fs x d S p e e d Z o n e Velocity K.E. ( Joules ) Braking ∆ x Residential 12 m/s (25mph) Main road 21 m/s (45mph) Freeway (legal) 32 m/s (65mph) Freeway (speeding) 40 m/s (85mph)

13 Example Fill out the following table for a car ( m = 1,600 kg ) Since K.E. depends on the square of the velocity, it increases dramatically as velocity increases. Example Fill out the following table for a car ( m = 1,600 kg ) at different velocities; μs = 0.9 and Fs = μs x FN and Wbreaks = Fs x d S p e e d Z o n e Velocity K.E. ( Joules ) Braking ∆ x Residential 12 m/s (25mph) 115,200 J 8.2 m Main road 21 m/s (45mph)  352,800 J 25 m Freeway (legal) 32 m/s (65mph) 819,200 J 58 m Freeway (speeding) 40 m/s (85mph) 1,280,000 J 91 m

14 = mass • gravity • height Unit = Joules (J)POTENTIAL ENERGY (P.E) Stored energy that is capable of doing work. Energy of position It can be electricity in a battery, energy in a compressed spring, a stretched rubber band, or an object that can fall. Gravitational P.E. – Energy associated with an object due to its height The GPE of an object at any given height can be found using the equation: P.E. = m • g • h = mass • gravity • height Unit = Joules (J)

15 When an object falls, the P.E. converts into K.E.As P.E. decreases, K.E. increases If there is no friction or air resistance, 100% of the P.E. will become K.E.

16 Gravitational PE does not depend on the path taken to get it there

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18 Elastic Potential EnergyEnergy stored in a spring that is stretched or compressed. The force on a spring is given by Hooke’s Law. F = k • d k = spring constant (or force constant) units of k = N/m The Elastic P.E. in the spring (the work it can do) is: P.E. = Favg • d ( Favg = Ftotal /2 ) P.E. = ½ kd2

19 ∆KE = ∆ PE CONSERVATION of ENERGYEnergy cannot be created or destroyed, It can only transform from one form to another. ∆KE = ∆ PE

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21 SI Units of Power = watt (W).In science ‘power’ means how fast work is being done. Power is the rate work is done (amount of work per time) Power = Work / Time P = W / ∆t OR Power = Force x distance / Time P = F x d / ∆t SI Units of Power = watt (W). (named after James Watt)

22 1 watt = 1 joule per second (1 J/sec)50 watt light bulb doing work at the rate of 50 joules per sec 1 kilowatt (kW) = 1000 watts. Electric power in your home is measured in kWh (kilowatt – hour). Unit of energy – not power.

23 1 horsepower can raise 330 pounds of coal 100 feet in a minute1 hp = 750 watts

24 POWER The rate of doing work: Example:The faster work is done, the more power it takes. Example: 1. An aircraft carrier catapult does 1.05 x 108 J of work in 2.2 s. a. How much power does it generate? b. How many horsepower is that? (1 hp = 750 W) 2. A dragster does 1.7 x 107 J of work in 4.0s

25 POWER Examples: The rate of doing work:The faster work is done, the more power it takes. Examples: 1. An aircraft carrier catapult does 1.05 x 108 J of work in 2.2 s. a. How much power does it generate? 4.8 x 107 watts b. How many horsepower is that? (1 hp = 750 W) 64,000 hp 2. A dragster does 1.7 x 107 J of work in 4.0s 4.3 x 10 6 watts 5,667 hp

26 Objectives: 1. Identify several forms of energy. 2. Calculate kinetic energy for an object. 3. Apply the work-kinetic energy theorem to solve problems. 4. Distinguish between kinetic and potential energy. 5. Classify different types of potential energy. 6. Calculate the potential energy associated with an object's position.