Week 4: Electric Motors & Mechanical Advantage

Grade 8 Science | Rosche | Kairos Academies

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The Phenomenon: The Tiny Motor Mystery

Anchoring Context & Focus Question

The Scene: A car jack motor sits in the palm of your hand โ€” smaller than your fist, drawing only 180 watts (about the same as two light bulbs). Yet it can lift a 1,500 kg car off the ground. That car weighs 14,700 Newtons โ€” roughly the weight of 1,500 textbooks stacked up.

Driving Question: How can a tiny electric motor lift a car?

Connection to Previous Weeks

W2: We learned that electric current creates magnetic fields and how to build an electromagnet. W3: We saw how electricity powers circuits. W4 (this week): Now we discover what happens when we use electromagnetic force to create motion โ€” and how gear systems multiply that force to do seemingly impossible work.

By the end of this week, you will be able to:

  • Explain how electric motors convert electrical energy to mechanical energy
  • Calculate motor efficiency using input and output power
  • Apply mechanical advantage principles to analyze force multiplication
  • Design a motor-and-gear system to meet specific lifting requirements

Vocabulary

Key Vocabulary (7 terms) — Practice Tool

Cognate Strategy: Many science words look similar in English and Spanish โ€” use your Spanish to learn science!

Term Spanish Definition
mechanical advantage ventaja mecรกnica Factor by which a machine multiplies force (MA = Output Force / Input Force)
gear ratio relaciรณn de engranajes Ratio of teeth on driven gear to driving gear; determines speed/torque trade-off
torque torque Rotational force โ€” the twisting power of a motor (measured in Newton-meters)
work trabajo Force times distance (W = F ร— d); measured in Joules
efficiency eficiencia Useful energy output divided by total energy input (always less than 100%)
pulley system sistema de poleas Wheels and ropes that change the direction or magnitude of force
energy energรญa Capacity to do work; can be transferred but not created or destroyed

Hook โ€“ The Tiny Motor Mystery

Compare motor size to car weight.

The Challenge: Do the Math

An electric car jack motor runs on 12 volts × 15 amps = 180 watts of power — about the same as two bright light bulbs. The car it lifts weighs 1,500 kg. Convert that to force:

Weight = mass × gravity = 1,500 kg × 9.8 m/s² = 14,700 N

The motor only produces a fraction of that force directly. So how does it lift the car? The answer is a GEAR / SCREW system. The motor uses mechanical advantage: it trades force for distance. The car rises very slowly (maybe 5 cm/minute), but the motor never needs to produce 14,700 N all at once. The screw mechanism multiplies the force for us.

Stop & Think

Think about when you use a bottle opener or scissors. Do those tools let you exert more force than you put in? How? What do you give up to get that extra force? Discuss with your partner before moving on.

COMPLETE THE HOOK FORM

Explore how small motors can produce large forces.

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Worked Example and Simulation โ€“ Calculating Mechanical Advantage in a Gear System

How to Calculate Motor Efficiency — Step by Step

The Scenario:

A motor lifts a 100 kg mass 2 meters high in 10 seconds. The motor draws 20 amps at 12 volts. What is the motor's efficiency?

Step 1: Calculate Input Power

Input power = the electrical power the motor draws from the source.

Pin = V × I = 12 V × 20 A = 240 W

Step 2: Calculate Output Work

Output work = the useful work done lifting the mass against gravity.

W = F × d = m × g × h = 100 kg × 9.8 m/s² × 2 m = 1,960 J

Step 3: Calculate Output Power

Output power = how fast the work was done.

Pout = Work ÷ Time = 1,960 J ÷ 10 s = 196 W

Step 4: Calculate Efficiency

Efficiency = what fraction of the input power became useful output power.

η = Pout ÷ Pin × 100% = 196 ÷ 240 × 100% = 82%

That means 82% of the electrical energy became motion. The other 18% became heat (friction in gears, resistance in wire).

Common Mistake: "Mechanical Advantage Creates Free Energy"

Wrong thinking: "If I use a 1:20 gear ratio, I get 20 times the force โ€” so I have 20 times the energy!"

Correct thinking: Mechanical advantage trades force for distance. If a gear multiplies force by 20, it reduces speed (and distance traveled) by 20. Work = Force × Distance. If force goes up ×20 and distance goes down ×20, the work stays the same. You never get more energy out than you put in โ€” you only change the form of the trade-off.

Simulation: Motor Efficiency Lab

PREDICT (before running the sim)

A motor converts electrical energy into mechanical energy, but some energy is always lost as heat. Predict: If you double the voltage, will the motor's efficiency increase, decrease, or stay the same? What about switching from a light load to a heavy load?

OBSERVE (while using the sim)

Test three voltages (3V, 6V, 12V) at each load level. Record: (1) Input power vs. output power at each setting. (2) How much energy is lost as heat? (3) Which motor type gives the highest efficiency for each load?

EXPLAIN (after collecting data)

Was your prediction correct? Use the terms input power, output power, and heat loss to explain: Why doesn't "more voltage = more efficiency"? What is the trade-off between torque and speed?

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Station 1 โ€“ Motor Efficiency Investigation

Analyze energy transfer in electric motors.

Key Formulas for Motor Efficiency

Formula Meaning Units
Pin = V × I Input power = Voltage × Current Watts (W)
Wout = m × 9.8 × h Output Work = mass × gravity × height Joules (J)
Pout = Wout ÷ t Output Power = Work ÷ Time Watts (W)
η = (Pout ÷ Pin) × 100% Efficiency (always < 100% due to friction/heat) Percent (%)

Data Table: Motor Efficiency Trials

Use the formulas above to fill in the missing columns in the form. The first trial is partially completed as a reference.

Trial Load (kg) Voltage (V) Current (A) Pin (W) Height (m) Time (s) Pout (W) Efficiency
1 2 12 1.5 18 1.0 5 3.9 22%
2 5 12 2.0 24 1.0 8 6.1 26%
3 10 12 3.5 42 1.0 14 7.0 17%

Hint: Why does efficiency drop for the heaviest load?

Heavier loads make gears and motors work harder, creating more friction. More friction means more energy lost as heat. That is why Trial 3 (heaviest load) has the lowest efficiency, even though the motor draws more current. Energy in > Energy out. Always.

COMPLETE THE STATION 1 FORM

Measure and calculate motor efficiency.

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Station 2 โ€“ Mechanical Advantage Analysis

Calculate mechanical advantage in motor systems.

Key Concept: Gear Ratio & Mechanical Advantage

A gear ratio tells you the relationship between input and output. A 1:5 gear ratio means for every 1 rotation of the input gear, the output gear turns 1/5 of a rotation — but with 5× the force. Work (Force × Distance) stays constant (minus friction losses).

Gear Ratio Input Force Output Force Input Distance Output Distance Example
1:1 10 N 10 N 10 cm 10 cm Direct drive
1:5 10 N 50 N 10 cm 2 cm Car jack
1:20 10 N 200 N 10 cm 0.5 cm Bicycle climbing gear
5:1 50 N 10 N 2 cm 10 cm Bicycle speed gear

Green = bigger  |  Red = smaller    Notice: force and distance always trade off. Work stays the same!

The Big Idea

Gear ratio multiplies force but reduces speed. A 1:20 gear gives you 20× the force, but your output moves 20× slower. That is why the car jack motor can lift a heavy car — it uses a large gear ratio to trade speed for force. The car rises slowly, but it rises!

COMPLETE THE STATION 2 FORM

Analyze gear ratios and pulley systems.

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Station 3 โ€“ Design a Lifting System

Apply electromagnetism and mechanical advantage to engineering.

COMPLETE THE STATION 3 FORM

Design a motor-powered system to lift a specified weight.

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Exit Ticket โ€“ Motor Systems Integration

Synthesize understanding of motors and mechanical advantage.

COMPLETE THE EXIT TICKET

Demonstrate mastery of motor and mechanical advantage concepts.

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Enrichment & Extension
Optional content if you finish early or want to go deeper.
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Week 4 Complete!

Next Week: Synthesis & Assessment โ€” bringing electricity and magnetism together!