logo

Balanced & Unbalanced Forces — Long Answer Questions (Class 9 Physics)

Medium Level (Application & Explanation)

Q1. Explain with examples how balanced forces act on a stationary object and on a moving object. Why does the state of motion not change under balanced forces?

Answer:

  • Balanced forces mean the vector sum of all forces on an object is zero. This results in no change in the object’s motion.
  • For a stationary object: A book on a table has gravity pulling it down and the normal force from the table pushing up. These forces are equal and opposite, so the net force is zero and the book remains at rest.
  • For a moving object: A car moving at constant speed on a straight road has the engine’s forward force balanced by friction and air resistance. Since net force is zero, the car continues moving with the same velocity.
  • Newton’s first law (inertia) explains this: an object keeps its state of motion unless an unbalanced force acts on it.
  • Thus, balanced forces do not change speed or direction — they preserve the current state of motion.

Q2. Describe static friction and kinetic friction. Give two real-life examples of each and explain why static friction is usually greater than kinetic friction.

Answer:

  • Static friction acts when surfaces are not moving relative to each other; it prevents motion up to a maximum value. Example 1: A heavy box on the floor that resists your starting push. Example 2: A parked car that does not slide on a slope because static friction holds it.
  • Kinetic friction acts when surfaces slide over each other; it opposes motion during sliding. Example 1: A sled sliding on snow. Example 2: A book sliding across a table.
  • Reason static > kinetic: Surfaces have microscopic irregularities that interlock more when at rest, requiring a larger force to start motion. Once sliding begins, these interlocks are partially broken, so less force (kinetic friction) opposes motion.

Q3. A child pushes a toy car with a constant force and it moves at constant velocity. Explain the forces acting and why the car does not accelerate. Include the role of friction in your explanation.

Answer:

  • The child applies a forward push on the toy car. Opposing this is friction (rolling or sliding) and air resistance acting backward.
  • If the car moves at constant velocity, the forward push equals the total opposing forces. The net force is therefore zero.
  • According to Newton’s first law, zero net force means no acceleration — the car maintains the same velocity.
  • Friction here converts some mechanical energy into heat, but as long as the push supplies energy equal to losses, speed stays constant.
  • If the push becomes slightly larger than friction, the car will accelerate; if smaller, it will slow down. Thus, balanced forces explain the steady motion.

Q4. Explain how friction can be both useful and harmful in everyday life. Provide three examples of usefulness and two examples of harm, and suggest ways to reduce harmful friction.

Answer:

  • Useful aspects:
    • Walking: Friction between shoes and ground gives traction so we don’t slip.
    • Brakes: Friction between brake pads and wheels stops vehicles safely.
    • Holding objects: Friction between our fingers and objects prevents them from slipping.
  • Harmful aspects:
    • Wear and tear: Friction in machine parts causes wear, reducing lifespan.
    • Energy loss: Friction converts useful energy into heat, lowering efficiency (e.g., engines).
  • Ways to reduce harmful friction:
    • Use lubricants (oil, grease) between moving parts.
    • Use ball bearings to change sliding friction to rolling friction.
    • Choose smoother materials and proper alignments to lessen contact resistance.
  • Thus friction is essential but must be controlled for efficiency and durability.

Q5. In a tug-of-war two teams pull with equal forces but the rope does not move. Later one team pulls harder and wins. Explain these situations using the concept of balanced and unbalanced forces, and state how net force changes.

Answer:

  • When both teams pull with equal force in opposite directions, the forces on the rope cancel out. The net force is zero — this is a case of balanced forces. So the rope does not move and remains in equilibrium.
  • When one team pulls harder, their force becomes greater than the opposing team’s force. The forces no longer cancel — resulting in a non-zero net force in the direction of the stronger team.
  • This unbalanced force causes the rope to accelerate toward the stronger side.
  • The magnitude of acceleration depends on the net force and the rope-plus-players’ effective mass: a larger net force produces a larger acceleration (Newton’s second law).

High Complexity (Analytical & Scenario-Based)

Q6. A 2 kg toy is pulled horizontally by a 6 N force across a table. The frictional force opposing motion is 2 N. Calculate the net force and acceleration. Explain each step and discuss how changing the friction would affect acceleration.

Answer:

  • Step 1 — Identify forces: Forward pull = 6 N, friction (opposite) = 2 N. Net force = forward − friction.
  • Step 2 — Compute net force: Net force = 6 N − 2 N = 4 N in the forward direction.
  • Step 3 — Use Newton’s second law: F = m × a → a = F / m = 4 N / 2 kg = 2 m/s².
  • Interpretation: The toy accelerates at 2 m/s² forward because of the unbalanced force of 4 N.
  • Effect of changing friction:
    • If friction increases, net force decreases, so acceleration decreases.
    • If friction equals 6 N, net force becomes zero and acceleration is zero (constant velocity if already moving).
    • If friction exceeds 6 N, net force reverses and the toy would decelerate or move backward.
  • This example shows how unbalanced forces determine acceleration.

Q7. Design an experiment to measure the coefficient of kinetic friction between a wooden block and a flat wooden surface using a spring balance. Describe procedure, observations, calculations and precautions.

Answer:

  • Aim: Find coefficient of kinetic friction (μk) between block and surface.
  • Apparatus: Wooden block, flat wooden plank, spring balance, stopwatch, measuring scale, weight set.
  • Procedure:
    • Place known mass block (m) on plank and attach spring balance horizontally.
    • Pull block at constant velocity; read the steady pull force Fk from spring balance (this equals kinetic friction).
    • Measure normal force = mg (if horizontal).
  • Calculation:
    • μk = Fk / N = Fk / (m × g), where g ≈ 9.8 m/s².
  • Observations:
    • Record Fk for different masses to verify μk remains roughly constant.
  • Precautions:
    • Ensure motion is constant speed (not accelerating).
    • Surface should be clean and uniform.
    • Pull horizontally to avoid vertical components changing normal force.
  • This method gives a practical value of μk and shows friction’s dependence on normal force.

Q8. A car moves at constant speed on a level road. Suddenly the engine stops and the car slows down to a halt. Explain all forces acting and why the car stops even though no single large opposing force seems present. Discuss energy conversion during this motion.

Answer:

  • When engine runs, driving force forward balances resistive forces (friction + air resistance), so net force = zero and speed is constant.
  • When engine stops, driving force disappears. Now only resistive forces act (friction in tires/axles and air drag) producing a net backward force.
  • This unbalanced force causes the car to decelerate and eventually stop.
  • Energy conversion: The car’s kinetic energy is converted into thermal energy by friction (brakes, tires, bearings) and into thermal energy of air through drag. Some energy may be dissipated as sound.
  • Even when resistive forces are not individually large, their combined effect provides sufficient net backward force to stop the car over time.
  • This illustrates how removal of forward force and ongoing resistive forces change motion.

Q9. A block rests on an inclined plane at 30° and does not slide. Explain the forces, resolve them along parallel and perpendicular directions, calculate the component of gravity along the plane for a 5 kg block, and determine the minimum static friction required to keep it stationary. (Take g = 9.8 m/s².)

Answer:

  • Forces acting: Weight (mg) acts downward, normal force (N) is perpendicular to plane, static friction fs acts up the plane to prevent sliding.
  • Resolve weight: Component parallel to plane = mg sinθ. Perpendicular component = mg cosθ.
  • For m = 5 kg, θ = 30°:
    • mg = 5 × 9.8 = 49 N.
    • Parallel component = 49 × sin30° = 49 × 0.5 = 24.5 N (down the plane).
    • Perpendicular component = 49 × cos30° ≈ 49 × 0.866 = 42.43 N.
  • Static friction requirement: To remain at rest, static friction must balance the parallel component, so minimum fs = 24.5 N (directed up the plane).
  • Maximum static friction available is μs × N; if μs × 42.43 N ≥ 24.5 N, block stays put. Otherwise it will slide.
  • This shows how components of gravity and normal force determine friction needs on inclines.

Q10. Two forces, 10 N east and 6 N at 60° north of east, act on a small object. Determine the resultant force vector (magnitude and direction). Explain how an unbalanced resultant leads to change in motion and describe qualitatively what would happen if the object was initially at rest.

Answer:

  • Resolve 6 N into components:
    • East component = 6 × cos60° = 6 × 0.5 = 3 N east.
    • North component = 6 × sin60° = 6 × 0.866 = 5.196 N north.
  • Combine east components: 10 N + 3 N = 13 N east. North component remains **5.196 N...

Back to Topics

Take the test