Force and Laws of Motion – Numerical Problems with Stepwise Solutions
CBSE Class 9 — Physics
Chapter 9: Force and Laws of Motion — 20 Numerical Problems with Stepwise Solutions
Content Bank — Important formulas & reminders
- Newton's 2nd law:
F_net = dp/dt, for constant massF = m·a. - Momentum:
p = m·v(unit: kg·m/s). - Impulse:
J = F·Δt = Δp(unit: N·s). - Friction:
f_s ≤ μ_s·N,f_k = μ_k·N. - Inclined plane: weight component along plane =
mg·sinθ; perpendicular =mg·cosθ. - Atwood's (two masses, frictionless pulley):
a = (m2 − m1)g/(m1 + m2), tensionT = m1(g + a)(care sign & directions). - Use SI units throughout. Clearly state assumptions (g = 9.8 m/s² unless stated otherwise).
Topic A — Basic F = ma & Units (Problems 1–4)
Problem 1
A net force of 18 N acts on a body of mass 3 kg. Find the acceleration of the body.
Step 1: Write Newton's second law:
F = m·a.Step 2: Substitute values:
a = F/m = 18 / 3.Step 3: Compute:
a = 6 m/s².Answer: 6 m/s².
Problem 2
A 5 kg object accelerates at 2 m/s². What is the net force acting on it?
Step 1: Use
F = m·a.Step 2:
F = 5 × 2 = 10 N.Answer: 10 N (in direction of acceleration).
Problem 3
A 12 N force produces acceleration 4 m/s² on a body. Find the mass.
Step 1: Rearrange
F = m·a → m = F / a.Step 2:
m = 12 / 4 = 3 kg.Answer: 3 kg.
Problem 4
A body initially at rest experiences a constant net force giving it an acceleration of 3 m/s² for 6 s. Find its final velocity.
Step 1: Use kinematic relation:
v = u + a·t, with u = 0.Step 2:
v = 0 + 3 × 6 = 18 m/s.Answer: 18 m/s.
Topic B — Newton's Laws Applications & Multiple Forces (Problems 5–8)
Problem 5
Three forces 10 N (right), 6 N (left) and 4 N (right) act on a particle along a straight line. If mass is 2 kg, find acceleration and direction.
Step 1: Choose right as positive. Net force = 10 − 6 + 4 = 8 N.
Step 2: Use
a = F_net / m = 8 / 2 = 4 m/s².Answer: 4 m/s² to the right.
Problem 6
A 20 N force acts on a 4 kg block to the right while friction of 6 N opposes motion. Find acceleration.
Step 1: Net force = applied − friction = 20 − 6 = 14 N (to the right).
Step 2:
a = 14 / 4 = 3.5 m/s².Answer: 3.5 m/s² to the right.
Problem 7
Two forces of 15 N and 25 N act on a 5 kg body at right angles to each other. Find magnitude of acceleration.
Step 1: Net force magnitude = √(15² + 25²) = √(225 + 625) = √850 ≈ 29.15 N.
Step 2:
a = F_net / m ≈ 29.15 / 5 ≈ 5.83 m/s².Answer: ≈ 5.83 m/s² (direction along resultant).
Problem 8
A box of mass 10 kg is pulled by two forces 40 N (right) and 30 N (right). If friction is 20 N, find acceleration.
Step 1: Total applied = 40 + 30 = 70 N. Net = 70 − 20 = 50 N.
Step 2:
a = 50 / 10 = 5 m/s².Answer: 5 m/s² to the right.
Topic C — Friction & Inclined Planes (Problems 9–12)
Problem 9
A 15 kg block is on a horizontal floor. Coefficient of kinetic friction μ_k = 0.2. A horizontal force 50 N is applied. Find acceleration (g = 9.8 m/s²).
Step 1: Normal N = mg = 15 × 9.8 = 147 N.
Step 2: Friction f_k = μ_k N = 0.2 × 147 = 29.4 N.
Step 3: Net force = 50 − 29.4 = 20.6 N.
a = 20.6 / 15 ≈ 1.373 m/s².Answer: ≈ 1.37 m/s² (to the direction of applied force).
Problem 10
A block of mass 4 kg is placed on an incline of angle 30°. If surface is frictionless, find acceleration down the plane (g = 9.8 m/s²).
Step 1: Component of weight along plane = mg·sinθ = 4 × 9.8 × sin30° = 39.2 × 0.5 = 19.6 N.
Step 2:
a = F_net / m = 19.6 / 4 = 4.9 m/s² down the plane.Answer: 4.9 m/s² down the plane.
Problem 11
A 10 kg block on a 37° incline (sin37° ≈ 0.6, cos37° ≈ 0.8) has coefficient of kinetic friction μ_k = 0.15. Find acceleration down the plane.
Step 1: Weight component along plane = mg·sinθ = 10×9.8×0.6 = 58.8 N.
Step 2: Normal = mg·cosθ = 10×9.8×0.8 = 78.4 N.
Step 3: Friction f_k = μ_k N = 0.15 × 78.4 = 11.76 N (opposes motion up the plane).
Step 4: Net force down = 58.8 − 11.76 = 47.04 N →
a = 47.04 / 10 = 4.704 m/s².Answer: ≈ 4.70 m/s² down the plane.
Problem 12
Find the minimum horizontal force needed to just start moving a 20 kg crate if μ_s = 0.4 (g = 9.8 m/s²).
Step 1: Normal N = mg = 20×9.8 = 196 N.
Step 2: Maximum static friction f_s(max) = μ_s N = 0.4×196 = 78.4 N.
Step 3: Minimum horizontal push > f_s(max); so required ≈ 78.4 N (slightly more to overcome).
Answer: ≈ 78.4 N (minimum to start motion).
Topic D — Connected Bodies & Pulleys (Problems 13–15)
Problem 13 (Atwood's machine)
Two masses m1 = 3 kg and m2 = 5 kg are connected over a frictionless, massless pulley. Find acceleration and tension (g = 9.8 m/s²).
Step 1: Let m2 descend, m1 ascend. Use
a = (m2 − m1)g / (m1 + m2).Step 2:
a = (5 − 3) × 9.8 / (5 + 3) = 2×9.8 / 8 = 19.6 / 8 = 2.45 m/s².Step 3: Tension using m1 equation:
T − m1 g = m1 a → T = m1(g + a) = 3(9.8 + 2.45) = 3×12.25 = 36.75 N.Answer: a = 2.45 m/s² (m2 down), T = 36.75 N.
Problem 14
A 6 kg mass hangs from a rope. The mass is pulled upward with acceleration 2 m/s². Find the tension in the rope (g = 9.8 m/s²).
Step 1: Along vertical upward positive:
T − mg = m a.Step 2:
T = m(g + a) = 6(9.8 + 2) = 6 × 11.8 = 70.8 N.Answer: 70.8 N.
Problem 15
Two blocks 4 kg and 2 kg are connected on a horizontal frictionless table and pulled by 18 N horizontal force on the 4 kg block. Find acceleration of system and tension in string between them.
Step 1: Total mass = 4 + 2 = 6 kg. Acceleration of system = F / M = 18 / 6 = 3 m/s².
Step 2: Tension in string (pulling 2 kg block) = m2 × a = 2 × 3 = 6 N.
Answer: a = 3 m/s², T = 6 N.
Topic E — Momentum, Impulse & Collisions (Problems 16–19)
Problem 16
A 0.5 kg ball moving at 8 m/s strikes a wall and rebounds in the opposite direction with speed 5 m/s. If impact time is 0.02 s, find average force on the ball (direction as sign).
Step 1: Initial momentum p_i = 0.5×8 = 4 kg·m/s (forward).
Step 2: Final momentum p_f = 0.5 × (−5) = −2.5 kg·m/s (opposite direction). Δp = p_f − p_i = −2.5 − 4 = −6.5 kg·m/s.
Step 3: Average force = Δp / Δt = −6.5 / 0.02 = −325 N. (Negative sign indicates force opposite initial motion.)
Answer: −325 N (i.e., 325 N opposite to initial direction).
Problem 17
A car of mass 1000 kg moving at 20 m/s collides and sticks to a stationary truck of mass 2000 kg. Find common velocity after collision and loss in kinetic energy.
Step 1: Conservation of momentum: (1000×20 + 2000×0) = (1000 + 2000) v → 20000 = 3000 v → v = 20000 / 3000 = 6.666... m/s.
Step 2: Initial KE = 0.5×1000×20² = 0.5×1000×400 = 200000 J. Final KE = 0.5×3000×(6.6667)² ≈ 0.5×3000×44.444 ≈ 66666.7 J.
Step 3: Loss = 200000 − 66666.7 ≈ 133333.3 J.
Answer: v ≈ 6.67 m/s; KE lost ≈ 1.33×10⁵ J.
Problem 18
A 0.25 kg tennis ball moving at 30 m/s is caught by a player who moves the glove backward so that the ball comes to rest in 0.5 s. What average force does the player feel?
Step 1: Δp = m(v_f − v_i) = 0.25(0 − 30) = −7.5 kg·m/s.
Step 2: Average force = Δp / Δt = −7.5 / 0.5 = −15 N (opposite to ball's initial motion).
Answer: −15 N (magnitude 15 N opposite to incoming ball).
Topic F — Braking, Stopping Distance & Practical (Problem 19–20)
Problem 19
A car moving at 25 m/s applies brakes and decelerates uniformly at 5 m/s². Calculate stopping distance.
Step 1: Use formula
v² = u² + 2 a s with final v = 0 and a = −5 m/s².Step 2:
0 = 25² + 2(−5) s → s = 25² / (2×5) = 625 / 10 = 62.5 m.Answer: 62.5 m.
Problem 20
A cyclist of mass 70 kg moving at 8 m/s applies brakes and stops in 4 s. Estimate average braking force (neglect rolling resistance).
Step 1: Δv = 0 − 8 = −8 m/s; Δt = 4 s → a = Δv/Δt = −2 m/s².
Step 2: Net force = m a = 70 × (−2) = −140 N. Magnitude 140 N opposite to motion.
Answer: Average braking force = 140 N (opposite to direction of motion).
Notes: All problems are aligned with NCERT Class 9 chapter on Force and Laws of Motion. Use SI units throughout. For exam practice, show each step (free-body diagrams where applicable), state assumptions (frictionless, g value) and check units and limiting cases. If you want, I can (a) include neat free-body diagrams for selected problems, (b) convert this page into a printable PDF, or (c) break the problems into printable worksheets by topic.