ENERGY Chapter 12 Section 3

Warm-up Name different types of energy and their daily uses.

What is Energy? Energy can be described as the ability to do work. Look at the picture below where do you believe energy is.

Potential Energy The energy that an object has because of the position, shape, or condition of the object. There are two types of potential energy: elastic and gravitational

Elastic Potential Energy The energy stored in any type of stretched or compressed elastic material. Name different objects that would have elastic potential energy.

Gravitational Potential Energy Energy that is stored in the gravitational field is known as gravitational potential energy. Depends on both mass and height. Let’s look at the equation used for gravitational potential energy

Gravitational Potential Energy Equation grav.PE= mass x free-fall acceleration x height Or in other terms PE= mgh PE= Potential energy, m= mass, g= free-fall acceleration, h= height

Example A 65 kg rock climber ascends a cliff. What is the climber’s gravitational potential energy at a point 35m above the base of the cliff?

How to solve this problem Step 1: State the equation you are going to use. PE=mgh

How to solve problem cont. Step 2: List the known and unknown values given in the equation Known values mass= 65 kg height= 35 m free-fall acceleration g= 9.8m/s² Unknown value- gravitational potential energy

How to solve the problem cont. Step 3: Plug in the values and solve. PE= (65 kg)(9.8m/s²)(35 m) PE= 2.2x 10⁴ kg x m²/s² = 2.2 x 10⁴ J

Practice Problems Try problem 1 and 2 on your own. Follow the steps of the example given above.

Warm up A book weighing 8.0 kg is on a table 75cm high. What is the gravitational potential energy of the book?

Warm-up Solved Step 1: State equation we are going to use PE = mgh Step 2: State known and unknown values Known values mass = 8.0kg g= 9.8m/s² h= 75cm Unknown values Potential Energy

Warm –up cont. Step 3: Plug in values and solve We have to convert 75cm to meters 75cm/ 100 = 0.75m PE = (8.0kg)(9.8m/s²)(.75m) PE = 58.8 J

Kinetic Energy The energy of a moving object due to the object’s motion Just like with potential energy the amount of kinetic energy is dependant on the objects mass as well as its speed

Kinetic Energy Equation Kinetic Energy = ½ mass x velocity² Or in other words KE = ½ mv² KE= kinetic energy m= mass v= velocity

Example Problem What is the kinetic energy of a 44 kg cheetah running at 31 m/s?

How to solve the problem Step 1: Write out the equation you are going to use KE= ½ m x v²

How to solve the problem cont. Step 2: Write out the known and unknown values m= 44kg v= 31 m/s Unknown KE= ?

How to solve the problem cont. Step 3: Plug in values and solve KE= ½ (44)(31)² KE= 2.1 x 10⁴ J

Mechanical Energy The amount of work an object can do because of the object’s kinetic and potential energies Remember all objects contain both kinetic and potential energy

Mechanical Energy can change forms Just like kinetic and potential energy mechanical energy can change forms as well. Changes from mechanical to nonmechanical energy due to the presence of friction and air resistance Nonmechanical energy is a special form of kinetic or potential energy

Conservation of Energy Section 12.4

Conservation of Energy Imagine you are riding the rollercoaster pictured below. Where is the energy coming from?

The Law of Conservation of Energy Energy can not be created or destroyed. What does that mean for the energy that is found around us.

Energy does not appear or disappear Energy works in systems. Two types of systems : closed and open

Closed System When the flow of energy into and out of a system is small enough to be ignored is called a closed system Based on the information above what is an example of a closed system.

Open System Occur when the energy exchange occurs within the space that surrounds them Why is a car an example of an open system

Efficiency A quantity usually expressed as a percentage that measures the ratio of useful work output to work input efficiency = useful work output/ work input Since this is expressed as a percentage we must multiply the answer by 100.

Efficiency Problem A sailor uses a rope and an old squeaky pulley to raise a sail that weighs 140 N. He finds that he must do 180 J of work on the rope in order to raise the sail by 1 m (doing 140 J of work on the sail). What is the efficiency of the pulley? Express your answer as a percentage.

Efficiency Problem Step 1: State the equation efficiency = useful work output/ work imput Step 2: State known and unknown values Known work input- 180 J Useful work output- 140 J Unknown values- efficiency percentage

Efficiency Problem Step 3: Solve 140/180= x 100 = 78 %

Practice Problems Try pg 407 # 1 and pg 408 # 7 for practice on your own