Math practice · Unit #06 · 7th grade · Forces

Tug-of-War Arithmetic

Forces have size AND direction — which makes them integers with an attitude. Right is positive, left is negative, and the net force is just the sum. Then: your first F = ma, with numbers that behave.

skills: integers net force F = ma ≈ 30 min
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Name Period Date
PART A

The net-force ledger

Tug-of-war: Team Amber pulls right (+), Team Iris pulls left (−). Net force = add them, signs included. Fill in the shaded cells — then write which way the rope moves (or "balanced").

Rope ledger (forces in newtons)
RoundAmber pulls (N)Iris pulls (N)Net force (N)Rope moves…
1+350−325
2+400−400
3+280−340
4+150 + 175−300

  1. In Round 2, is anything happening? The rope doesn't move — but is anyone pulling? Explain "balanced forces" in one sentence using the word zero.

  2. A cart already rolling right gets a net force of 0 N for 5 seconds. What does its motion do — speed up, slow down, or keep rolling the same? (First law preview.)

PART B

First contact: F = ma

F = m × a force (N) = mass (kg) × acceleration (m/s²)
Cover the thing you don't know; multiply or divide the other two. Show every setup.

  1. A 3 kg cart is pushed with 12 N. What is its acceleration?

  2. What force gives a 2 kg cart an acceleration of 5 m/s²?

  3. A 100 N push makes a scooter accelerate at 4 m/s². What is the scooter's mass?

  4. Same push (24 N) on two carts: cart X is 2 kg, cart Y is 6 kg. Find both accelerations. What did the extra mass do to the acceleration?

PART C

Your weight on other worlds

Weight = mass × gravitational field (g). Your mass never changes — your weight is a local decision, made by whichever planet is pulling. Use a 50 kg student.

Gravitational field strength (NASA planetary data)
Worldg (N/kg)Weight of 50 kg student (N)% of Earth weight
Earth9.8490100%
Moon1.6
Mars3.7
Jupiter24.8

  1. On which world could you lift the heaviest backpack? Why — use the word mass or weight correctly in your answer.

  2. Your friend says "I'd weigh nothing on the Moon because there's no gravity." Use a number from the table to correct them — kindly.

PART D

Net force, then acceleration

A force system has more than one push. Add the directions first, then use the net force in a = F ÷ m. Friction counts as a force too.

  1. A sled is pulled right with 40 N. Two forces pull left: 12 N and friction at 8 N. Find the net force and direction.

  2. The sled has a mass of 5 kg. Use the net force from Question 1 to find its acceleration.

  3. If friction grows, what happens to the net force and acceleration? Explain using one signed number.

Answer key — teachers

Part A

R1: +25 N → right · R2: 0 N → balanced (doesn't move) · R3: −60 N → left · R4: +325 − 300 = +25 N → right.

  1. Balanced forces cancel to zero — everyone is pulling, but the net is zero, so nothing changes.
  2. Keeps rolling the same. Zero net force = no CHANGE in motion (not no motion).

Part B

  1. a = F ÷ m = 12 ÷ 3 = 4 m/s²
  2. F = m × a = 2 × 5 = 10 N
  3. m = F ÷ a = 100 ÷ 4 = 25 kg
  4. X: 24 ÷ 2 = 12 m/s²; Y: 24 ÷ 6 = 4 m/s². Triple the mass → one-third the acceleration (more mass resists more).

Part C

Moon: 80 N (≈16%) · Mars: 185 N (≈38%) · Jupiter: 1,240 N (≈253%).

  1. The Moon — backpack weight there is ~16% of Earth's; its mass is unchanged but it weighs far less.
  2. Moon weight = 80 N, not 0 N — the Moon's pull is weaker, not absent. (Hammer-and-feather moment optional.)

Part D

  1. Net force = 40 − 12 − 8 = 20 N right.
  2. Acceleration = 20 ÷ 5 = 4 m/s² right.
  3. More friction makes the left-side force larger, so the net force and rightward acceleration become smaller. For example, 12 N of friction would give 16 N right.