1. Foundations of Physics
CPO Science
Foundations of Physics
Chapter 9
Unit 4, Chapter 10

2. Unit 4: Energy and Momentum
Chapter 10 Work and Energy
10.1 Machines and Mechanical Advantage
10.2 Work
10.3 Energy and Conservation of Energy

3. Chapter 10 Objectives
Calculate the mechanical advantage for a lever or rope and pulleys.
Calculate the work done in joules for situations involving force and distance.
Give examples of energy and transformation of energy from one form to another.
Calculate potential and kinetic energy.
Apply the law of energy conservation to systems involving potential and kinetic energy.

4. 10.1 Machines and Mechanical Advantage
Key Question:
How do simple machines work?

5. 10.1 Machines
In physics the term “simple machine” means a machine that uses only the forces directly applied and accomplishes its task with a single motion.

6. 10.1 Machines
The best way to analyze what a machine does is to think about the machine in terms of input and output.

7. 10.1 Mechanical Advantage
Mechanical advantage is the ratio of output force to input force.
For a typical automotive jack the mechanical advantage is 30 or more.
A force of 100 newtons (22.5 pounds) applied to the input arm of the jack produces an output force of 3,000 newtons (675 pounds)— enough to lift one corner of an automobile.

8. 10.1 Mechanical Advantage 𝑴𝑨= 𝑭 𝒐 𝑭 𝒊 Output force (N) Mechanical
𝑴𝑨= 𝑭 𝒐 𝑭 𝒊
Output force (N)
Mechanical
advantage
Input force (N)

9. 10.1 Mechanical Advantage of a Lever
𝑴𝑨 𝒍𝒆𝒗𝒆𝒓 = 𝑳 𝒊 𝑳 𝒐
Length of input arm
(m)
Mechanical
advantage
Length of output arm
(m)

11. 10.1 Calculate position
Where should the fulcrum of a lever be placed so one person weighing 700 N can lift the edge of a stone block with a mass of 500 kg?
The lever is a steel bar three meters long.
Assume a person can produce an input force equal to their own weight.
Assume that the output force of the lever must equal half the weight of the block to lift one edge.
1) You are asked to figure out the location of the fulcrum.
2) You are given the input force, length of the lever, and the mass to be lifted.
3) The weight of an object is equal to its mass times 9.8 N/kg. The mechanical advantage
of a lever is the ratio of length of the input arm divided by length of the output arm.
4) Solve the problem:
Output force = Fo = (1/2)(500 kg)(9.8 N/kg) = 2,450 N
The required mechanical advantage is: MA = Fo/Fi = (2,450 N)/(700 N) = 3.5.
The mechanical advantage of a lever is the ratio of lengths: MA = Li/Lo = 3.5.
From this equation you know that Li = 3.5 Lo.
The total length of the lever is 3 meters; that means Li + Lo = 3 m.
Substitute for the length of the input arm to solve for the length of the output arm.
(3.5 Lo) + Lo = 3 m Lo = 3 m
Lo = 0.67 m, the fulcrum should be placed 0.67 meters from the edge of the block.

13. 10.1 Wheels, gears, and rotating machines
Axles and wheels provide advantages.
Friction occurs where the wheel and axle touch or where the wheel touches a surface.
Rolling motion creates less wearing away of material compared with two surfaces sliding over each other.
With gears the trade-off is made between torque and rotation speed.
An output gear will turn with more torque when it rotates slower than the input gear.

14. 10.1 Ramps and Screws
Ramps reduce input force by increasing the distance over which the input force needs to act.
A screw is a simple machine that turns rotating motion into linear motion.
A thread wraps around a screw at an angle, like the angle of a ramp.

15. 10.2 Work
Key Question:
What are the consequences of multiplying forces in machines?

16. 10.2 Work In physics, work has a very specific meaning.
In physics, work represents a measurable change in the energy of a system, caused by a force.

17. 10.2 Work
If you push a box with a force of one newton for a distance of one meter, you have done exactly one joule of work.

18. 10.2 Work (force is parallel to distance)
Force (N)
Work (joules)
W = F x d
Distance (m)

19. 10.2 Work (force at angle to distance)
Force (N)
Work (joules)
W = Fd cos (q)
Angle
Distance (m)

21. 10.2 Work done against gravity
Mass (g)
Height object raised (m)
Work (joules)
W = mgh
Gravity (m/sec2)

22. 10.3 Why the path doesn't matter

23. 10.3 Calculate work
A crane lifts a steel beam with a mass of 1,500 kg.
Calculate how much work is done against gravity if the beam is lifted 50 meters in the air.
How much time does it take to lift the beam if the motor of the crane can do 10,000 joules of work per second?
1) You are asked for the work and the time it takes to do the work.
2) You are given mass, height, and the work done per second.
3) Use the formula for work done against gravity, W = mgh.
4) Solve:
W = (1,500 kg)(9.8 N/kg)(50 m) = 735,000 J
At a rate of 10,000 J/sec, it takes: 735,000 ÷ 10,000 = 73.5 seconds to lift the beam.

25. 10.3 Energy and Conservation of Energy
Energy is the ability to make things change.
A system that has energy has the ability to do work.
Energy is measured in the same units as work because energy is transferred during the action of work.

26. 10.3 Forms of Energy
Mechanical energy is the energy possessed by an object due to its motion or its position.
Radiant energy includes light, microwaves, radio waves, x-rays, and other forms of electromagnetic waves.
Nuclear energy is released when heavy atoms in matter are split up or light atoms are put together.
The electrical energy we use is derived from other sources of energy.

28. 10.3 Potential Energy Ep = mgh Mass (kg) Potential Energy (joules)
Height (m)
Acceleration
of gravity (m/sec2)

29. 10.3 Potential Energy
A cart with a mass of 102 kg is pushed up a ramp.
The top of the ramp is 4 meters higher than the bottom.
How much potential energy is gained by the cart?
If an average student can do 50 joules of work each second, how much time does it take to get up the ramp?
1) You are asked for the potential energy and time.
2) You are given mass, height, and the work done per second.
3) Use the formula for potential energy Ep = mgh.
4) Solve:
Ep = (102 kg)(9.8 N/kg)(4 m) = 3,998 J
At a rate of 50 J/sec, it takes:
3,998 ÷ 50 = 80 seconds to push the cart up the ramp.

30. 10.3 Kinetic Energy Energy of motion is called kinetic energy.
The kinetic energy of a moving object depends on two things: mass and speed.
Kinetic energy is proportional to mass.

31. 10.3 Kinetic Energy
Mathematically, kinetic energy increases as the square of speed.
If the speed of an object doubles, its kinetic energy increases four times. (mass is constant)

32. Ek = ½ mv2 10.3 Kinetic Energy Mass (kg) Kinetic Energy Speed (m/sec)
(joules)
Ek = ½ mv2
Speed (m/sec)

33. 10.3 Kinetic Energy
Kinetic energy becomes important in calculating braking distance.

34. 10.3 Calculate Kinetic Energy
A car with a mass of 1,300 kg is going straight ahead at a speed of 30 m/sec (67 mph).
The brakes can supply a force of 9,500 N.
Calculate:
a) The kinetic energy of the car.
b) The distance it takes to stop.
1) You are asked for the kinetic energy and stopping distance.
2) You are given mass, speed, and the force from the brakes.
3) Kinetic energy Ek = 1/2 mv2 Work, W = Fd
4) Solve:
Ek = (1/2)(1,300 kg)(30 m/sec)2 = 585,000 J
To stop the car, the kinetic energy must be reduced to zero by work done by the brakes.
585,000 J = (9,500 N) × d d = 62 meters

35. 10.3 Law of Conservation of Energy
As energy takes different forms and changes things by doing work, nature keeps perfect track of the total.
No new energy is created and no existing energy is destroyed.

36. 10.3 Energy and Conservation of Energy
Key Question:
How is motion on a track related to energy?

37. Application: Hydroelectric Power