Forces — Chapter review · Physical Science: Matter, Forces & Energy, Grade 8

Model balanced forces, gravity, friction, electric and magnetic interactions, and fluid pressure.

Review the key ideas

Balanced & Unbalanced Forces

Add forces in the same direction, subtract forces in opposite directions; a net force of zero means balanced (no change in motion), otherwise the object accelerates.

Interaction Pairs & Force Diagrams

Interaction partners act on different objects; acceleration depends on the net force on each object separately. The cart’s net force is 8−3=5 N rightward.

Gravity & Friction

Weight = mass × g (W = mg) and depends on gravity, while mass never changes; friction always opposes motion between touching surfaces.

Electric & Magnetic Forces

Electric and magnetic forces can act through space; controlled tests reveal what changes their strength.

Pressure & Buoyant Forces

Fluids exert pressure. A steadily floating object has balanced buoyant force and weight.

Problem 1

Can balanced forces act on an object that is moving?
Answer and explanation

Yes. With zero net force its velocity does not change. It can remain at rest or continue moving at constant speed in a straight line.

Yes. With zero net force its velocity does not change. It can remain at rest or continue moving at constant speed in a straight line.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 2

Could the top cart be moving right?
Answer and explanation

Yes, at constant velocity in the stated simple model. Zero net force does not require zero velocity.

Yes, at constant velocity in the stated simple model. Zero net force does not require zero velocity.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 3

Visual-model chapter review: A 3 kg cart receives 14 N right and 5 N left while initially moving left. Find net force and acceleration and describe the initial speed change.
Answer and explanation

Net 9 N right; acceleration 9/3 = 3 m/s² right. Its initial leftward motion slows because acceleration opposes velocity.

Subtract opposing forces on the same receiver.

Compare acceleration direction with the supplied velocity.

Net 9 N right; acceleration 9/3 = 3 m/s² right. Its initial leftward motion slows because acceleration opposes velocity.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 4

Why do the two forces not cancel in the cart diagram?
Answer and explanation

Only one acts on the cart. Adding a force on the person to the forces on the cart mixes receivers and gives an invalid net force.

Only one acts on the cart. Adding a force on the person to the forces on the cart mixes receivers and gives an invalid net force.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 5

With 3 N leftward friction, what is the cart’s horizontal net force?
Answer and explanation

8−3=5 N rightward. The cart-on-hand partner remains 8 N leftward; it is not the cart’s net force.

8−3=5 N rightward. The cart-on-hand partner remains 8 N leftward; it is not the cart’s net force.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 6

Visual-model chapter review: A cart accelerates right at 1.5 m/s² with mass 4 kg. The recorded push is 10 N right. What horizontal resistance would make the introductory force model agree, and where does the cart-on-hand partner force belong?
Answer and explanation

Required net is 4×1.5 = 6 N right, leaving 4 N resistance left. The cart-on-hand partner belongs on the hand’s diagram, not the cart’s.

Use measured acceleration to find net force.

Keep the third-law partner on its separate receiver.

Required net is 4×1.5 = 6 N right, leaving 4 N resistance left. The cart-on-hand partner belongs on the hand’s diagram, not the cart’s.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 7

Why does greater weight not automatically give greater ideal free-fall acceleration?
Answer and explanation

Greater mass also means greater resistance to acceleration. Since gravitational force is mg, force divided by mass gives g for each object when drag is neglected.

Greater mass also means greater resistance to acceleration. Since gravitational force is mg, force divided by mass gives g for each object when drag is neglected.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 8

Why does the object retain its mass on the Moon?
Answer and explanation

Moving locations changes the gravitational field and weight, not the stated amount of matter.

Moving locations changes the gravitational field and weight, not the stated amount of matter.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 9

Use the table to construct a gravitational claim about mass and separation. Then explain why a feather and stone can fall differently in air without proving different ideal free-fall acceleration.
Hints

Use the illustrated study titled Gravity: masses, distance and free fall. Keep the observations, model and conditions separate.

Answer and explanation

The force rises from 8 to 16 relative units when one mass doubles at fixed separation, and falls to 2 when separation doubles at the reference masses. These model comparisons support attractive, mass- and distance-dependent interaction under the stated conditions. Feather/stone differences in air also involve drag and inertia; an ideal negligible-drag comparison does not depend on mass in the same way. Separate receivers for the interaction pair and name the model limit.

The force rises from 8 to 16 relative units when one mass doubles at fixed separation, and falls to 2 when separation doubles at the reference masses. These model comparisons support attractive, mass- and distance-dependent interaction under the stated conditions. Feather/stone differences in air also involve drag and inertia; an ideal negligible-drag comparison does not depend on mass in the same way. Separate receivers for the interaction pair and name the model limit.

Response criteria

  • Compare one changed factor at a time.
  • Explain attraction and appropriate force receivers.
  • Distinguish gravity-only motion from significant drag.

Problem 10

In the no-drag throw with initial horizontal velocity 6 m/s, what are horizontal velocity, vertical velocity and vertical acceleration at the top? Does gravity stop acting?
Answer and explanation

Horizontal velocity 6 m/s; vertical velocity 0; vertical acceleration −10 m/s². Gravity continues acting downward.

Separate the two velocity components.

Distinguish zero vertical velocity from zero acceleration.

Response criteria

  • State all three values with units.
  • Keep gravity downward.
  • Avoid claiming zero total velocity.

Problem 11

Visual-model chapter review: For a no-drag ball launched upward at 8 m/s and horizontally at 3 m/s, use g = 10 m/s². At 0.8 s state both velocity components and acceleration, then predict horizontal range when it returns to release height.
Answer and explanation

At the top after 0.8 s: horizontal velocity 3 m/s, vertical velocity 0, vertical acceleration −10 m/s². Total flight is 1.6 s, so range is 4.8 m under the same-height ideal model.

Vertical velocity loses 10 m/s each second, reaching zero after 0.8 s.

Symmetric ideal same-height return takes twice that time; multiply by horizontal speed.

At the top after 0.8 s: horizontal velocity 3 m/s, vertical velocity 0, vertical acceleration −10 m/s². Total flight is 1.6 s, so range is 4.8 m under the same-height ideal model.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 12

Does attraction prove both objects have opposite net charges?
Answer and explanation

No. A charged object can attract a neutral object through polarization. Repulsion in a suitable simple test can be stronger evidence of like net charges.

No. A charged object can attract a neutral object through polarization. Repulsion in a suitable simple test can be stronger evidence of like net charges.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 13

Why not change the test magnet size between positions?
Answer and explanation

That would change another cause of force, confounding the intended distance comparison.

That would change another cause of force, confounding the intended distance comparison.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 14

Use the coil records to ask an investigable question about each changed factor. Then draw a repelling-magnet energy model and contrast the work needed for separating an attractive pair.
Hints

Use the illustrated study titled Field strength and stored energy. Keep the observations, model and conditions separate.

Answer and explanation

At fixed 0.2 A, changing turns from 10 to 20 raises the illustrative lift count 3→5; at fixed 20 turns, current 0.2→0.3 A changes it 5→7. These proxy outcomes support questions and a bounded trend, not a direct calibrated force law. Moving repelling magnets closer requires input work; separating attracting magnets can require input work instead. Energy is associated with the pair’s arrangement and can transfer into later motion.

At fixed 0.2 A, changing turns from 10 to 20 raises the illustrative lift count 3→5; at fixed 20 turns, current 0.2→0.3 A changes it 5→7. These proxy outcomes support questions and a bounded trend, not a direct calibrated force law. Moving repelling magnets closer requires input work; separating attracting magnets can require input work instead. Energy is associated with the pair’s arrangement and can transfer into later motion.

Response criteria

  • Isolate turns/current using the proper records.
  • Distinguish proxy measurement from force.
  • Model potential energy, orientation and work.

Problem 15

An electric-field arrow at a location points upward. Predict force direction on a positive test charge and on a negative test charge. Explain why both cannot simply follow the same arrow.
Answer and explanation

Positive: upward. Negative: downward. Field direction uses the positive-test-charge convention.

Identify the definition of electric-field direction.

Reverse the force direction for a negative test charge.

Response criteria

  • Name both directions.
  • Explain the sign convention.
  • Distinguish field from a path.

Problem 16

Use the closed-loop and cut-magnet figures. Give the field direction outside and inside a bar magnet, then predict the poles on each cut piece.
Answer and explanation

Outside N to S; inside S to N. Each cut piece has both N and S poles.

Trace a complete loop.

Identify both poles on each smaller magnet.

Response criteria

  • Give both directions.
  • Keep both poles on each piece.
  • Distinguish magnetic poles from electric charge.

Problem 17

Visual-model chapter review: A small negative charge is placed where an electric-field arrow points left. Predict the electric force direction. Then explain why an attracted neutral paper piece need not acquire a net charge.
Answer and explanation

The negative charge feels force right, opposite the positive-test-charge convention. A neutral paper piece can polarize while keeping balanced total positive and negative charge.

Use the field convention before choosing the test-charge force.

Distinguish charge redistribution from transfer that changes net charge.

The negative charge feels force right, opposite the positive-test-charge convention. A neutral paper piece can polarize while keeping balanced total positive and negative charge.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.

Problem 18

At equal depths in the same stationary liquid, does a wider container necessarily give greater pressure?
Answer and explanation

No. With the same surface pressure and liquid density, pressure at a given depth does not depend simply on container width.

No. With the same surface pressure and liquid density, pressure at a given depth does not depend simply on container width.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 19

Which displacement must be smaller in the ideal lift?
Answer and explanation

The larger-area output piston moves a shorter distance when displaced fluid volumes match.

The larger-area output piston moves a shorter distance when displaced fluid volumes match.

Response criteria

  • Use the supplied conditions and relevant evidence.
  • Explain the scientific mechanism or calculation.
  • State a meaningful limitation or testable next step.

Problem 20

Floating, sinking and being held are different casesFloating, sinking and being held are different casesFloating steadilyReleased heavy blockHeld by a stringWeight 3 N downBuoyancy 3 N upNet force = 0 NWeight 4 N downBuoyancy 2 N upNet = 2 N downNet force = 0 NThese are separate cases. Arrows use 25 drawing units per newton.Predict acceleration from the net force; an object can still have velocity.Weight:4 N downBuoyancy:2 N upString:2 N upBlue horizontal lines mark the water surface.
Interpret the different support conditions in this diagram, then apply the same method to the changed force values in the question. The labels are the illustrated examples, not the numerical answer to the new case.
Visual-model chapter review: A submerged object is held still by a string. Its weight is 9 N and buoyant force 6 N. What upward string force balances it, and why would drawing only equal buoyancy and weight misdescribe the case?
Answer and explanation

The string supplies 3 N upward. Buoyancy alone is smaller than weight; the third force is needed to make the held object’s net force zero.

List all forces on the object.

Use 6+T−9 = 0 without calling the held state unsupported floating.

The string supplies 3 N upward. Buoyancy alone is smaller than weight; the third force is needed to make the held object’s net force zero.

Response criteria

  • Use the named labels, directions or supplied quantities accurately.
  • Connect the visual evidence to the scientific mechanism or calculation.
  • State the relevant model conditions or limitation.
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