1. Work, Power, & Energy Homework: Read pages 257 – 260
Answer questions p 261 # 1, 2, 3
P 262 #5, 6, 7, 8

2. Recall:
A change in an object’s momentum is related to the amount of time a force acts on an object.
Now we will look at the effect of a force acting on an object over a distance.

3. Work
The product of a force exerted on an object and the displacement of the object as a result of the force

4. When is work done on an object?
1. a force is applied And when 2. object moves in the direction that the force is applied

5. Ex) work is done when an object is lifted against gravity
The greater the mass of the object, the more work is done
- If an object does not move in the direction of the force, no work is done

6. Two categories
1. Work done against another force Ex) archer stretches a bowstring (work done against elastic forces in bow) 2. Work done to change the speed of an object Ex) bringing an automobile up to speed on a highway **Simplest case: force is constant and motion takes place in a straight line

7. Formula:
Work = Force x distance
W = F d
Units: 1 Joule = 1 N x 1 m

8. The third floor of a house is 8 m above street level
The third floor of a house is 8 m above street level. How much work is needed to move a 150 kg refrigerator to the third floor?

9. Determine the work done by a 45
Determine the work done by a 45.0 N force in pulling a suitcase at an angle of 50.0⁰ for a distance of 75.0 m.

10. Power Definition: the rate at which work is done
Formula: Power = work done / time interval
Unit: Watt = Joule / second
1 watt of power is expended when 1 Joule of work is done in one second

11. Power
High power engines don’t necessarily do more work, they just do the same amount of work in less time.
An engine that has 2x the power of a smaller engine will do the same amount of work in half the time.

12. A small electric motor is used to lift a 0
A small electric motor is used to lift a 0.50-kilogram mass at constant speed. If the mass is lifted a vertical distance of 1.5 meters in 5.0 seconds, what is the average power developed by the motor?

13. Energy – the ability to do work
Types of energy: mechanical thermal electric electromagnetic nuclear chemical

14. Mechanical Energy Kinetic Energy energy of motion KE = ½ m v2
Potential Energy Stored energy or Energy of position g PE = m g h

15. Work –Energy Theorem
Moving objects can do work on another object that it comes into contact with
A moving object exerts a force on a second object and moves it a distance
Therefore, an object in motion has the ability to do work and can be said to have energy (KE)

16. Calculations using the work-energy theorem
The work done on an object is equal to its change in KE
W = D KE
W = KEf – KEi
W = ½ m vf2 – ½ m vi2
if positive work is done on an object, the KE increases by the work done
if negative work is done, the KE of the object decreases by the work done (force acts in the opposite direction of motion)

17. If a 1490 kg car at rest is pushed a distance of 25 m until it reaches a speed of 2.0 m/s.
What was the car’s change in kinetic energy?
DKE = 2980 J
What was the work done on the car by the individual pushing the car?
W = 2980 J
What was the average force supplied by the pusher over the 25 m?
F = N

18. Kingda Ka accelerates riders from rest to 57.2 m/s over 100 m.
What is the change in KE of an 80.0 kg student, while the ride accelerates him?
How much work does the seat do on the rider?
What is the average force of the seat on the rider?

19. Types of Potential Energy
Gravitational PE
2. Elastic PE

20. Gravitational Potential Energy (gPE)
an object can have PE because of position relative to Earth
object has the ability to do work as a result of it falling
in order to lift a mass, m, vertically a force at least equal to its weight must be exerted on it
the work done lifting the object vertically is equal to the PE of the object
W = DPE
W = Force (Weight) x distance (height)
W = (m x g) x (h)
the gPE is dependent on the vertical height measured from some reference point (floor, sea level)
the work an object can do when it falls does not depend on the path taken, but only on the vertical height (free fall; down an incline)

21. Elastic Potential (springs, rubber bands, bungee cords)
1. PE associated with elastic substances
Ex) when an object is stretched or compressed it has the potential to do work when returning to its original (equilibrium) length
The change in length of stretch or compression (x) is proportional to the force exerted (F) to do the stretching or compressing
Fs = kx k = spring constant
k measures the stiffness of the elastic substance
The spring itself exerts a restoring force in the opposite direction of the displacement.

22. Elastic Potential
Stretched or compressed elastic objects have potential energy as a result of their ability to do exert a restoring force. The PE stored in the spring can be determined from the following: PEs = ½ k x2

23. Energy Conversions
Recall: The work done by all forces acting on an object is equal to the total change in Kinetic and Potential energy on the object W = DKE + DPE

24. Energy Transformations:
When dealing with mechanical energy we typically deal with Kinetic and Potential
Energy can be transformed from one form to another
Ex) a falling object decreases its PE, yet its velocity increases constantly, therefore its KE increases
Water falling at Niagara Falls has PE, while it falls the KE is used to turn turbines, which are used to convert mechanical energy to electrical
Work is done when energy is transferred from one object to another.

25. Conservation of Energy
**Whenever energy is transformed, no energy is lost or gained in the process.
Consider a falling object, neglecting air resistance, of course:
Before being dropped the PE is mgh (with reference to some point)
as it falls the height of the object decreases,
therefore PE must decrease
also as it falls, the object accelerates to ground,
so velocity is increasing
therefore KE must also be increasing as given by ½ mv2

26. Conservation of Energy (continued)
Just before our object hits the ground, all of the PE is converted to KE
So KEbottom = PEtop

27. Law of Conservation of Energy
The total energy is neither increased nor decreased in any process.
Energy can be transformed form one form to another, and transferred from one object to another, but the total energy remains constant.
KE + PE = constant OR
Total E at pt 1 = Total E at pt 2

28. The diagram below shows a 0
The diagram below shows a 0.1-kilogram apple attached to a branch of a tree 2 meters above a spring on the ground below.
The apple falls and hits the spring, compressing it 0.1 meter from its rest position. If all of the gravitational potential energy of the apple on the tree is transferred to the spring when it is compressed, what is the spring constant of this spring?
Answer : k = N/m

29. Mechanical Energy (with friction)
In real world problems as potential energy is converted to kinetic energy, and vice versa, friction can do work to impede the energy conversion. Some mechanical energy can be converted to heat energy (Q) during the conversion. When friction is considered we use the following equation: ET = PE + KE + Q

30. A roller coaster car has a potential energy of 450,000 J and a kinetic energy of 120,000 J at point A along its track. At the lowest point in the ride the potential energy is zero. And 50,000 J of work have been done by friction after it leaves point A. What is the kinetic energy of the car at its lowest point?
ET = PE + KE + Q 450,000 J + 120,000 J = 0 J + KE + 50,000 J 520,000 J = KE

31. KEtop + PEtop = KEbottom + PEbottom
A 420 N child sits on a swing that hangs 0.40 m above the ground. Her mother pulls the swing back and releases her when the seat is 1.5 m above the ground.
How fast is she moving when the swing passes through its lowest point?
KEtop + PEtop = KEbottom + PEbottom
½ mv2 + mgDh = ½ mv2 + mgh
(420 N)( m) = ½ (42.8 kg)(v2)
v = 4.6 m/s

32. (420 N)(1.5m – 0.4 m) = ½ (42.8 kg)(3.5 m/s)2 + Q
b. If she is moving through the lowest point at 3.5 m/s, how much energy was lost to friction?
ET = PE + KE + Q
(mgDh)top = 0 J + ½ mv2 + Q
(420 N)(1.5m – 0.4 m) = ½ (42.8 kg)(3.5 m/s)2 + Q
200 J = Q

33. mghtop = mghbottom + ½ mv2bottom + Q
A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires 3.13 × 105 joules of kinetic energy as it descends to the bottom of a 115 m hill in 5.3 seconds.
Calculate the energy “lost” do to friction as the roller coaster went down the hill.
ET = PE + KE + Q
mghtop = mghbottom + ½ mv2bottom + Q
(290 kg)(9.81m/s2)(115 m) = 0 J x 105 J + Q
14,200 J = Q

34. KE = ½ mv2 3.13 x 105 J = ½ (290 kg)(v2) 46.5 m/s = v
Calculate the speed of the roller coaster car at the bottom of the hill.
KE = ½ mv x 105 J = ½ (290 kg)(v2) 46.5 m/s = v

35. a = Dv/t a = (46.5 m/s – 0 m/s) / 5.3 s a = 8.8 m/s
Calculate the magnitude of the average acceleration of the roller coaster car as it descends
a = Dv/t a = (46.5 m/s – 0 m/s) / 5.3 s a = 8.8 m/s