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.

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 = 3.75 x 109 J

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.

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)

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 = = 1840 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 = 1840 J = 147 N cos ϴ x d

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.

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

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 = 12675 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 ) = 0.0468 J or 4.7 x 10 -2 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

5-2 continuation: WORK – ENERGY THEOREM   When 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 = 45 and KE = 352 J Find ∆X (d) = ? Formula: Wnet = ∆ KE F ● ∆X = ∆ KE Wnet = F ● ∆X ∆X = ∆ KE / F = 7.8 m   

5-2 continuation: WORK – ENERGY THEOREM   When 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

How far will it Skid?

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)

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

= 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)

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.

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

Elastic Potential Energy Energy 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

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

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)

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.

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

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

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

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.