Energy and its TransformationsSimple Machines,

Work & Power

Chapter 15

Examples

Simple Machines and Work

• A simple machine is a device used to make work easier.

• It is used to multiply forces or change the direction of the force.

• No machine is 100% efficient.• Energy is always lost to friction.• The person must do more work, but easier

work, i.e. less force.• A lesser force is applied over a greater

distance.

Definitions & Equation

• Effort Force, FE: Force used by person, N

• Effort Distance, ΔdE: Distance person’s force is applied over, m.

• Resistance Force, FR: Force that comes from machine, or weight of object, N

• Resistance Distance, ΔdR: Distance object is moved, m.

• FE x ΔdE = FR x ΔdR

Mechanical Advantage

• Mechanical Advantage is a measure of how much easier the task has become.

• Ideal Mechanical Advantage – no friction:• IMA = ΔdE = Effort arm length ΔdR Resistance arm length• Actual Mechanical Advantage – with friction.• AMA = FR

FE

Levers

• The fulcrum is a pivot that can act to change the direction and magnitude of the force.

• Depending on force direction and position of fulcrum:– 1st class– 2nd class– 3rd class

1st Class Lever

• 1st Class Lever: The fulcrum, or pivot, is located between the 2 forces.

• E.g. lever, seesaw, teeter-totter, scissors.

2nd Class Lever

• 2nd Class Lever: The fulcrum is located on 1 side of both forces.

• E.g. wheel barrow, nutcracker

3rd Class Lever

• 3rd Class Lever: The effort force is between the resistance force and the fulcrum.

• E.g. tweezers, hockey stick

Variations of the Lever• The pulley is a variation of the lever.

• IMA = the number of supporting ropes.

• IMA = ΔdE

ΔdR

• AMA = FR

FE

Wheel and Axle

• The wheel and axle acts like a 2nd class lever. E.g. doorknob, taps, steering wheel

• IMA = radiusE AMA = FR

radiusR FE

The Inclined Plane

• The inclined plane trades distance for force by reducing the force needed to work against gravity.

• E.g. stairs, ramps, screw

Inclined Plane Formula

• IMA = length of slope = ΔdE

height of slope ΔdR

• AMA = FR = mg

FE FE

Activity

• A 1st class lever has a AMA of 4. How much force is needed to lift 5 kg?

• AMA = FR = mg

FE FE

• 4 = 5kg x 9.81 m/s2

FE

• FE = 12.25 N

Exam QuestionA workman uses a pulley to lift a 50 kg sack of potatoes by pulling downwards on a rope with a force of 550 N.

x

What is the acceleration of the sack? A)

1.0 m/s2

B)

2.1 m/s2

C)

4.5 m/s2

D)

11 m/s2

Measuring Work

• Work is defined as the energy that comes from applying a force over a certain distance.

• W = F Δd = mad (horizontal)

• = magd (against gravity)

• Work is in Joules, J

• Force is in Newtons, N

• Distance is in metres, m

Activity

• E.g How much work is done by a boy pushing a car with a force of 800 N over a distance of 200m?

• W = F d

• = 800 N x 200 m

• = 160 000 J = 160 kJ

• Do page 330, Q. 1-4

Exam QuestionA 200 g brick falls from a wall 4.0 metres above the ground. It hits the ground with a velocity of 8.5 m/s.

4.0 m

How much work did gravity do on the brick?

A)

8.0 J

B)

7.2 J

C)

3.4 J

D)

1.7 J

Exam QuestionA sled has a mass of 10 kg. A child pulls the sled a distance of 20 metres with a force of 10.0 N at an angle of 35° with respect to the horizontal. During this motion, a force of friction of 4.0 N acts in the opposite direction of the motion.

35

10.0 N

How much work is done on the sled by the child over the distance of 20 metres?

A)

1.6 102 J

B)

1.1 102 J

C)

8.4 101 J

D)

3.5 101 J

Efficiency

• The IMA is always greater than the AMA.

• The MA must be greater than 1.

• % Efficiency = Work output x 100 = AMA

Work input IMA

• The maximum efficiency is 100%.

• It is a measure of what energy is lost to friction, vibration, and other factors.

Power

• Power is defined as the rate at which work is being done.

• P = W

• Δt

• Work is in Joules,

• Time is in seconds

• Power is in Watts

Activity

• What is the power of a bulldozer that does 55000J of work in 1.1s?

• P = W = 55000J = 50000 Watts

Δt 1.1s

• If 100000 J of energy was expended by the bulldozer, what is its efficiency?

• Do page 334, Q 1-5

Exam Question

A horse is hitched up to a buggy with a mass of 500 kg including the people inside. Disregard the effects of friction. Starting from rest, the horse exerts a horizontal force of 300 newtons on the buggy over a distance of 30 metres. What is the average power that the horse develops over the first 30 metres?

A)

9.0 102 W

B)

9.0 103 W

C)

4.5 105 W

D)

4.5 106 W

Summary

• Work done on object equals the applied force times the displacement of the object in the direction of the force.

• W = FΔd• The units are Joules, J, or Nm.• Power is the rate at which work is done

per unit of time.• P = W/Δt• The units are Watts, W, or J/s

Summary

• The Efficiency of a machine is given by the formula, Efficiency = Work output x 100

Work input• A machine is supposed to reduce the force

required by a person to do a physical task.• Ideal Mechanical Advantage does not take

into account friction.• IMA = effort distance • resistance distance

Summary

• Actual Mechanical Advantage, AMA, takes friction into account.

• AMA = Resistance Force

Effort Force

Exam Question

A car with a mass of 1000 kg and moving at a speed of 30 m/s comes to rest over a distance of 100 metres. What is the force of friction (acting on the wheels of the car) which causes the car to stop?

A)

3 000 N

B)

3 500 N

C)

4 000 N

D)

4 500 N

Activity

• Do Page 338, Q. 1-6