Forces & Newton's Laws

A resultant force of 12 N acts on a body of mass 3 kg. Find the magnitude of its acceleration.

• A36 m/s²
• B4 m/s²✓
• C9 m/s²
• D0.25 m/s²

Answer: By Newton's second law F = ma, so a = F/m = 12/3 = 4 m/s². Option A multiplies instead of dividing, and D inverts the fraction.

A box has mass 7 kg. Taking g = 9.8 m/s², find the weight of the box.

• A68.6 N✓
• B7 N
• C70 N
• D0.71 N

Answer: Weight is W = mg = 7 × 9.8 = 68.6 N. Option C uses g = 10, option B confuses mass with weight, and D divides by g.

Two horizontal forces act on a particle of mass 2 kg lying on a smooth surface: 15 N to the right and 9 N to the left. Find the magnitude of the particle's acceleration.

• A12 m/s²
• B7.5 m/s²
• C3 m/s²✓
• D4.5 m/s²

Answer: The resultant force is 15 − 9 = 6 N to the right. Then a = F/m = 6/2 = 3 m/s². Option A divides the sum 24 N by 2, and B divides only the 15 N.

Two particles of mass 3 kg and 5 kg are connected by a light inextensible string passing over a smooth fixed pulley, and released from rest. Taking g = 9.8 m/s², find the acceleration of the system.

• A9.8 m/s²
• B19.6 m/s²
• C4.9 m/s²
• D2.45 m/s²✓

Answer: For the system, (5 − 3)g = (5 + 3)a, so a = 2 × 9.8 / 8 = 2.45 m/s². Option C wrongly divides by the mass difference; A and B ignore the masses.

Two particles of mass 3 kg and 5 kg hang from a light inextensible string over a smooth fixed pulley and are released from rest; the system accelerates at 2.45 m/s². Taking g = 9.8 m/s², find the tension in the string.

• A49 N
• B36.75 N✓
• C29.4 N
• D19.6 N

Answer: Consider the 3 kg mass, which accelerates upward: T − 3g = 3a, so T = 3(9.8 + 2.45) = 3 × 12.25 = 36.75 N. (Checking with the 5 kg mass: 5(9.8 − 2.45) = 36.75 N.) Option C is 3g and D is wrong sign handling.

A block of mass 4 kg rests on a rough horizontal surface with coefficient of friction μ = 0.3. Taking g = 9.8 m/s², find the maximum friction force before the block begins to slip.

• A1.2 N
• B39.2 N
• C11.76 N✓
• D12 N

Answer: In limiting equilibrium the friction is F = μR = μmg = 0.3 × 4 × 9.8 = 11.76 N (the normal reaction R = mg on a horizontal surface). Option D uses g = 10, A forgets the weight, and B omits μ.

A box of weight 50 N rests in equilibrium on a smooth slope inclined at 30° to the horizontal, held by a force P acting up the slope (parallel to the slope). Find the magnitude of P.

• A50 N
• B25 N✓
• C43.3 N
• D100 N

Answer: Resolving parallel to the smooth slope, P balances the component of weight down the slope: P = 50 sin 30° = 50 × 0.5 = 25 N. Option C uses 50 cos 30°, the component perpendicular to the slope.