# Energy Physics 11. Types of Energy  Name as many different types of energy as you can. Think of an example to go with each.

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Energy Physics 11

Types of Energy  Name as many different types of energy as you can. Think of an example to go with each.

Key Terms  Look up in your glossary (or in Chapter 6 of the text book, specifically 6.2) the following key terms:  Energy  Kinetic Energy  Potential Energy  Elastic Potential Energy  Gravitational Potential Energy  Mechanical Energy

Types of Energy  There are 2 main classifications of energy:  1) Potential Energy – The energy stored in a body or system as a consequence of its position, shape or form. Example: An object being held up has potential energy because of its position (gravitational potential energy). Example: A compressed spring has potential energy (potential to spring open).  2) Kinetic Energy – The energy of motion Example: When you walk across the classroom you have kinetic energy.

Quick “Lab”  Go around the room to inspect the different toys.  Decide what type(s) of energy makes them work (move, make noise, etc) and write ALL the energy types down on the sheet provided.

What is energy?  The measure of a system’s ability to do work.  http://www.youtube.com/watch?v=y UpiV2I_IRI http://www.youtube.com/watch?v=y UpiV2I_IRI  Minutes 13 onward  19:30 onward

Work and Energy Relationship  There is not much difference between work and energy.  In order to do work, an object must have energy.  In order to have energy, an object must have work done on it.

Energy Formulae  Kinetic Energy = E k = KE = ½ mv 2  Gravitational Potential Energy = E g = PE = mgh  Units: Joules (J)  NOTE: h = height of the object measured from the reference level (measured in metres).

What is the effect of doing work on an object?  You can give an object more kinetic energy by doing more work on it.  W = ΔKE = change in kinetic energy  W = KE f – KE i = KE final – KE initial  W = ½ mv f 2 - ½ mv i 2

Work-Kinetic Energy Theorem  W = ΔKE  The work done is equal to the change in kinetic energy of an object.

Example 1  A 145g tennis ball is thrown at a speed of 25m/s.  A) What is the ball’s kinetic energy?  B) How much work was done to reach this speed assuming the ball started from rest.

Answers  A) 45 J  B) W = ΔKE = 45 J – 0J  ** It was thrown from rest so kinetic energy initial is zero!!!

Ex 2: Work on an moving object A 2kg object is moving at 10 m/s when a force is applied to it accelerating it to 20 m/s over a distance of 5m. What is the work done on the object?

Check your understanding  A 1500. kg car is travelling at a speed of 50.0 km/h when the driver applies the brakes. The car comes to a stop after travelling 15 m.  a) How much work was done by friction in stopping the car?  b) What was the magnitude of the average frictional force applied to the car?  c) If the same car were travelling at 100. km/h when the driver applied the brakes and the car experienced the same frictional force, how far would the car travel before coming to a stop?

Worksheet

Ex 3: Potential Energy and Work What is the work done on a 12kg object to raise it from the ground to a height of 1.5m?

Potential Energy and Work What is the work done on a 12kg object to raise it from the ground to a height of 1.5m?

Ex 4:  A 1000kg car moves from point A to point B and then point C. The vertical distance between A and B is 10.m and between A and C is 15 m.  A) What is the PE at B and C relative to A?  B) What is the ΔPE (ΔPE = PE f – PE i ) when it goes from B to C?  C) Repeat a) and b) but take the reference level at C (switch all letters).

Answers  A) B: 98100 J  98000J, C: -147150J  -150 000J  B) a decrease of 245250 J  250000J  C) A: 150 000J B: 245250 J  250 000J Difference from A to B: increase of 100000J

Assignment on Energy – Kinetic and Potential

Comprehension Check  A truck pushes a car by exerting a horizontal force of 500. N on it. A frictional force of 300. N opposes the car’s motion as it moves 4.0m.  A) Calculate the work done on the car by the truck.  B) Calculate the work done on the car by friction.  C) Calculate the work done on the car overall (net work).

Answers  A) W = Fd = 500 x 4 = 2000 N = 2.00 x 10 3 J  B) W = Fd = -300 x 4 = -1200. J  C) W net = 2000 – 1200 = 800J

Comprehension Check  Calculate the work done by a horse that exerts an applied force of 100. N on a sleigh, if the harness makes an angle of 30’ with the ground and the sleigh moves 30.m across a flat, level ice surface (ie, no friction).

Answer  W = Fd cosΘ = (100)(30)cos(30) = 2.6 x 10 3 J

Comprehension Check  A 50. kg crate is pulled 40. m along a horizontal floor by a constant force exerted by a person (100. N) which acts at an angle of 37’. The floor is rough and exerts a force of friction of 50.N. Determine the work done by EACH FORCE acting on the crate, and the net work done on the crate.  DRAW A DIAGRAM!!!

 W Fg = FdcosΘ  Work is 0J as the force is perpendicular to gravitational force.  W FN = 0J (same reason as above)  W Fapp = Fdcos Θ =(100)(40)cos37’ = 3195J  S.F.  3200 J  W Ff = Fd = -50(40) = -2000 J  -2.0 x 10 3 J  Wnet = 3200 – 2000 = 1200 J

Comprehension Check  Mrs. Evans is holding a 2.4kg textbook at a height of 3.4m above the floor.  a) What is the type of energy (potential or kinetic)? How do you know?  b) How much energy is there (use your equation)?  c) What is the velocity of the book at this point (ie, velocity initial)?  d) If Mrs. Evans drops the book, what is the final velocity assuming she doesn’t throw it (use your kinematics equations!)?  e) If Mrs. Evans drops the book as in d), what is the type of energy when the book hits the floor?  f) How much of this energy is there when it touches the floor?  g) Is there any time when there are both kinds of energy? If so, when? Explain.

Answers  A) Potential: It is not moving, it has the potential to move (fall).  B) PE = mgh = 2.4x9.81x3.4 = 80.J  C) v = 0 (at rest, not thrown)  D) vf 2 = vi 2 + 2ad = 2(9.81)(3.4) = 66.708 Vf = 8.2m/s [down] E)KE = ½ mv 2 = ½ (2.4)(8.2) 2 = 80.J F)80.J G)When the object is falling, there is both PE and KE. When it falls 1.7m, there is equal PE and KE. Before this point (higher than 1.7m) there is more PE. After this (lower than 1.7m) there is more KE.

Work-Energy Theorem  “The net work done on an object is equal to its change in energy"  If the object is experiencing KE:  if Wnet is +ve, KE increases (moves in direction of force or speeds up)  if Wnet is -ve, KE decreases (moves in direction of friction or slows down)  If the object is experiencing PE:  if Wnet is +ve, PE increases (is lifted)  if Wnet is -ve, PE decreases (is lowered)

Total Energy and Work-Energy Theorem  The total energy of an object is the kinetic energy added to the potential energy.  As an object is dropped, the kinetic energy changes to potential energy until there is 0 PE and only KE.  E T = PE + KE

Law of Conservation of Energy  Within a closed, isolated system, energy can change form, but the total amount of energy is constant  Closed - no objects enter or leave the system.  Isolated - no net external force is exerted on it.  Energy cannot be created or destroyed  E 1 = E 2  KE 1 + PE 1 = KE 2 + PE 2

Examples  Book Drop  Collision into a spring  Car coasting up a hill h v v x hv v

Example 1:  A heavy object is dropped. If this object reaches the floor at a speed of 3.2 m/s from what height was it dropped?

Answer  Etop = Ebottom  KE + PE = KE + PE  0 + PE = KE + 0  mgh = ½ mv 2  NOTE: The masses will cancel!  gh = ½ v 2  9.81h = ½ (3.2) 2  h = 0.52 m

Example 2:  A heavy box slides down a frictionless incline. The incline has a slope of 30° and the length of the incline is 12m. If the box starts from rest at the top of the incline what is the speed at the bottom?

Example 3:  A 4.0 x 10 4 kg roller coaster starts from rest at point A. Neglecting friction, calculate its potential energy relative to the ground, its kinetic energy and its speed at points B,C and D in the illustration above.

Page 287  Questions 1, 2, 3, 4, 6, 8  A 15.0 kg box slides down an incline. If the box starts from rest at the top of the incline and has a speed of 6.0m/s at the bottom, how much work was done to overcome friction?  NOTE: The incline is 5.0m high (vertically) and the incline that the box goes down is 8.0m long (hypotenuse).  Remember: W = ΔE

Try This …  A skier is gliding along with a speed of 2.00m/s at the top of a ski hill. The hill is 40.0m high. The skier slides down the icy (frictionless) hill.  A) What will the skier’s speed be at a height of 25.0m? B) At what height will the skier have a speed of 10.0m/s?  HINT: Use similar triangles!

 Known: vi = 2.00m/s  h1 = 40.0m  h2 = 25.0m

REMEMBER…  W = ΔE  That E can be potential or kinetic energy!