2.  A force does work on an object if it causes the object to move. Work is always done on an object and causes a change in the object. Work is not energy itself but is the transfer of mechanical energy.  There are two essential elements of work: a force acting on an object and the object moving over some distance.

3.  Formula: W = ∆d  Units: Nm = J (joule) ◦ One joule is equal to one Newton of force on an object over one metre.  Note: work is a scalar quantity.

4.  Ex. You push your desk across the 25m room and exert a force of 105N. How much work was done to get the desk across the room?  Known: = 105N, d = 25m  Unknown: W  Solution: W = ∆d = (105)(25) = 2625J = 2600J

5.  1). When the force does not cause motion. This may happen when you are pushing on a wall or an object that is too heavy.  2). When there is uniform motion and no force. If a curling stone is sliding down the ice at a constant velocity with no external force acting on it then there is no work.  3). When the force is perpendicular to the motion. If you are carrying a book while walking you are putting a force on the book that is perpendicular to you walk.

6.  This is when the force and displacement are not parallel and are pointing in the same direction.  Formula: W = FCosθΔd : Note: θ is the angle between the force and direction.

7.  Ex. 2. You are pulling a wagon at a 40.° with a force of 35N over a distance of 110m. How much work was done?  Known: θ=40., F = 35, d = 110  Unknown: W  Solution: W = FCosθΔd = (35)(Cos40)(110) = 2850J = 2900J

8.  Positive work is when energy is added to an object.  Negative work is when energy is removed from an object.  For example, when you kick a soccer ball you transfer energy from your foot to the ball. So, the work from foot-to-ball is positive and the work from ball-to-foot is negative.  For negative work W = -FΔd

9.  Pg. 221 #1-3  Pg. 225 #4-10  Pg. 235 #14-18

10.  This is the energy of motion.  It is calculated by using half the mass of an object and the square of it’s velocity.  Formula: E k = (1/2)mv 2  Units: J (joule)

11.  Ex. A 0.43kg baseball is thrown at 44.7 m/s. What is the kinetic energy of the ball?  Known: m = 0.43, v = 44.7  Unknown: E k  Solution: E k = (1/2)mv 2 = (1/2)(0.43)(44.7) 2 = 429.6J = 430J

12.  This is the relationship between the work done on an object and the resulting kinetic energy.  It says that work equals the change in kinetic energy. Look back and notice that both work and kinetic energy have the same units.  Formula: W = E k2 – E k1 : Δd = ((1/2)mv 2 2 ) – ((1/2)mv 1 2 )  Units: J (joule)

13.  Ex. A 0.105kg hockey puck is shot 30m down the length of the ice. To do this a force of 15N was added.  A). How much work was done on the puck?  B). What was the velocity at the point of release?  Known: F = 15N, m = 0.105kg, d = 30m, v 1 =0m/s.  Unknown: W and v 2

14.  Solution: Δd = ((1/2)mv 2 2 ) – ((1/2)mv 1 2 )  A). W = (15)(30) = 450 J  B). 450 = ((1/2)(0.105)(v 2 2 )) – ((1/2)(0.105)(0) v 2 2 = 450/0.0525 v 2 = 92.6 m/s

15.  Pg. 238 #19-21  Pg. 245-246 #22-26

16.  There are 2 types of potential energy: gravitational and elastic.

17.  This is exactly as it sounds, it is the stored energy due to the effects of gravity. When you hold an object in the air it has potential energy based on what effect gravity will have when you drop it.

18.  Formula: E g = mgΔh  Units: N·m = J  Ex. You about to drop a water balloon on an unsuspecting, walking target below. If the water balloon has a mass of 1.2 kg and is 4.5 m above the target what is the potential energy?  Known: m = 1.2, g = 9.81, h = 4.5  Unknown: E g  Solution: E g = mgΔh = (1.2)(9.81)(4.5) = 52.97 J = 53 J

19.  Formula: W = E g2 – E g1  Ex. You get your favorite kite stuck in a tree. You (81kg) need to climb up to get it. If it takes 3650J of work to get to it, how high was the kite in the tree?  Known: W=3650, m=81, g=9.81, E g1 =0  Unknown: h

20.  Solution: W = E g2 – E g1 W = mgΔh – 0 3650 = (81)(9.81)(h) 4.59m = h 4.6m = h

21.  If an object can return to its original condition it is considered to be elastic. Since the object can change form, one of those forms must store energy. And this is called the elastic potential energy.  Think of a spring or elastic band.

22.  An elastic object can be compressed or stretched and when released will exert a force that will bring them back to their original form. This is the restoring force and it acts in the opposite direction to the stretch of compression.  The applied force is a result of the spring constant and the amount of extension or compression on the object.

23.  Formula: F = -kx  Note: k = spring constant (measured in N/m) and x = amount of extension or compression on the object (measured in m).  Units: (N/m)(m) = N

24.  Ex. If it takes 85N to hold a compressed spring and the length of the compression was 0.55 m, what is the spring constant?  Known: F = 85.0, x = 0.550.  Unknown: k  Solution: F = -kx 85 = -(k)(0.55) -154.6 = k -155 N/m = k

25.  The potential elastic energy in a perfectly elastic object equals the product half of the spring constant and square of the amount of compression or extension.  Formula: E e = (1/2)kx 2  Units: J (joule)

26.  Ex. An elastic band with a spring constant of 65 N/m is stretched to 45cm, what is it’s elastic potential energy?  Known: x = 0.45, k = 65  Unknown: E e  Solution: E e = (1/2)kx 2 = (1/2)(65)(0.45 2 ) = 6.58J = 6.6J

27.  Pg. 254 #30-34  Pg. 258# 35-37  Pg. 261 # 38-40

28.  Power is the rate at which work is done. This means it is the rate as which energy is transferred.  Formula: P = W/∆t or P = E/∆t  Units: J/s = W (watt)  Note: one watt is one joule of energy transferred per second. For example, a 100W light transfers 100 joules per second. 1 horsepower (hp) is equal to 746 W.

29.  Ex. A water pump can do 350J of work in 7.0 seconds. What is the power of the pump?  Known: W = 350J, t =7s  Unknown: P  Solution: P = W/∆t = (350)/(7) = 50. W

30.  This is the description of how much input energy a machine converts into output. We know that no machine is 100% efficient, there is always some energy “lost”. (We know energy is never really lost just transformed, usually into heat).

31.  Formula: Efficiency = E o /E I ·100 Efficiency = W o /W I ·100  Units: Efficiency has no units because it is a ratio.  Ex. If a water pump takes 475J of energy to do 340.J of useful work, how efficient is the pump?  Known: E o = 340., E I = 475  Unknown: Efficiency  Solution: Efficiency = E o /E I ·100 = (340/475)(100) = 71.6%

32.  Pg. 266#41-43  Pg. 270-271 # 44-50