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. Multiplying instead of dividing gives 12 × 3 = 36 m/s²; inverting the fraction to 3/12 gives 0.25 m/s²; and subtracting the mass from the force gives 9 m/s².

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. Taking g = 10 gives 70 N; quoting 7 N confuses the mass in kilograms with the weight in newtons; and dividing by g rather than multiplying gives 7/9.8 = 0.71 N.

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. Adding the forces to 24 N and dividing by 2 gives 12 m/s²; using only the 15 N gives 7.5 m/s²; and using only the 9 N gives 4.5 m/s².

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. Dividing the net force by the mass difference instead of the total mass gives 9.8 m/s², and the values 19.6 and 4.9 are simply 2g and g/2, which do not follow from the equation of motion.

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.) The value 29.4 N is 3g, the weight of the smaller mass, and 19.6 N is the net driving force (5 − 3)g on the system, not the tension.

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). Taking g = 10 gives 12 N; multiplying μ by the mass alone and omitting g gives 0.3 × 4 = 1.2 N; and leaving out μ gives the normal reaction itself, 4 × 9.8 = 39.2 N.

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. Using 50 cos 30° = 43.3 N gives the component perpendicular to the slope, which the normal reaction balances instead.

A resultant force acts on a mass of 4 kg, giving it an acceleration of 3 m s⁻². Find the force.

• A1.33 N
• B12 N✓
• C0.75 N
• D7 N

Answer: Newton's second law: F = ma = 4 × 3 = 12 N. The units confirm it — kg × m s⁻² is the newton, so the two quantities must be multiplied.

A resultant force acts on a mass of 12 kg, giving it an acceleration of 2.5 m s⁻². Find the force.

• A4.8 N
• B30 N✓
• C0.208 N
• D14.5 N

Answer: Newton's second law: F = ma = 12 × 2.5 = 30 N. The units confirm it — kg × m s⁻² is the newton, so the two quantities must be multiplied.

A resultant force acts on a mass of 0.5 kg, giving it an acceleration of 8 m s⁻². Find the force.

• A8.5 N
• B0.0625 N
• C16 N
• D4 N✓

Answer: Newton's second law: F = ma = 0.5 × 8 = 4 N. The units confirm it — kg × m s⁻² is the newton, so the two quantities must be multiplied.

Find the weight of a 5 kg mass, taking g = 9.8 m s⁻².

• A24.5 N
• B5 N
• C0.510 N
• D49 N✓

Answer: Weight is a force: W = mg = 5 × 9.8 = 49 N. Mass is measured in kilograms and weight in newtons — quoting 5 N confuses the two.

Find the weight of an 8 kg mass, taking g = 9.8 m s⁻².

• A78.4 N✓
• B8 N
• C0.816 N
• D39.2 N

Answer: Weight is a force: W = mg = 8 × 9.8 = 78.4 N. Dividing by g instead gives 0.816, and halving g gives 39.2. Mass is measured in kilograms and weight in newtons — quoting 8 N confuses the two.

Find the weight of a 20 kg mass, taking g = 9.8 m s⁻².

• A20 N
• B196 N✓
• C2.04 N
• D98 N

Answer: Weight is a force: W = mg = 20 × 9.8 = 196 N. Mass is measured in kilograms and weight in newtons — quoting 20 N confuses the two.

A 3 kg block on a rough horizontal surface is pulled by a 20 N horizontal force against 8 N of friction. Find its acceleration.

• A6.67 m s⁻²
• B4 m s⁻²✓
• C9.33 m s⁻²
• D2.67 m s⁻²

Answer: Work with the RESULTANT force: 20 − 8 = 12 N. Then a = F/m = 12/3 = 4 m s⁻². Using the 20 N applied force alone and ignoring friction gives 6.67, which is the usual error.

A book rests on a table. Which pair are Newton's third law partners?

• AThe table's weight and the floor's reaction
• BThe book's weight and the table's normal reaction on the book
• CThe book's weight and the book's mass
• DThe book's push down on the table and the table's push up on the book✓

Answer: Third-law pairs act on DIFFERENT bodies, are the same type of force and are equal and opposite. Weight and normal reaction both act on the book, so they are not a pair — they happen to balance here, which is Newton's first law, not his third.

Two particles of mass 5 kg and 3 kg are connected by a light inextensible string over a smooth pulley. Which statement is correct?

• AEach particle has its own separate acceleration
• BThe heavier particle accelerates faster than the lighter one
• CThe tension equals the weight of the heavier particle
• DBoth accelerate at the same magnitude, and the string tension is the same throughout✓

Answer: An inextensible string forces both particles to move at the same rate, so the magnitudes of their accelerations are equal. A light string over a smooth pulley has the same tension along its whole length. If the tension equalled the heavier weight, that particle could not accelerate at all.

State Newton's first law.

• AThe momentum of any closed system always remains constant over time
• BThe acceleration of a body is directly proportional to the resultant force applied to it
• CEvery action has an equal and opposite reaction acting on a different body
• DA body stays at rest or at constant velocity unless a resultant force acts✓

Answer: The first law is about equilibrium: no resultant force means no change in motion. The second law (acceleration proportional to resultant force) and the third (equal and opposite forces on different bodies) are both offered — knowing which is which is worth a mark on its own.

A uniform beam AB, of length 5 m and mass 30 kg, rests horizontally on supports at A and B. A child of mass 40 kg sits on the beam 1.5 m from A. Taking g = 9.8 m s⁻², find the reaction at A.

• A264.6 N
• B421.4 N✓
• C274.4 N
• D343.0 N

Answer: Moments about B: R_A × 5 = 30g × 2.5 + 40g × 3.5 = 75g + 140g = 215g, so R_A = 43g = 421.4 N. (Check: R_B = 70g − 43g = 27g = 264.6 N.) The value 264.6 N is the reaction at B, and 274.4 N ignores the weight of the beam.

A uniform ladder of length 4 m and mass 20 kg rests with its foot on rough horizontal ground and its top against a smooth vertical wall. The ladder makes an angle of 60° with the ground and is on the point of slipping. Find the coefficient of friction between the ladder and the ground, to 3 significant figures.

• Aμ = 0.866
• Bμ = 0.577
• Cμ = 0.289✓
• Dμ = 0.144

Answer: Resolving: the ground's normal reaction R = 20g and friction F equals the wall's reaction S. Moments about the foot: S × 4 sin 60° = 20g × 2 cos 60°, so S = 10g/tan 60° = 10g/√3. Then μ = F/R = (10g/√3)/(20g) = 1/(2√3) = 0.289. Swapping sine and cosine in the moment equation gives tan 60°/2 = 0.866, and placing the weight at the top of the ladder instead of its midpoint gives 1/√3 = 0.577.

A non-uniform rod AB, of length 2 m and mass 8 kg, hangs horizontally from two vertical strings, one at each end. The tension in the string at A is three times the tension in the string at B. Find the distance of the rod's centre of mass from A.

• A1.5 m
• B0.5 m✓
• C0.75 m
• D0.25 m

Answer: Resolving: T_A + T_B = 8g with T_A = 3T_B, so T_B = 2g and T_A = 6g. Moments about A: T_B × 2 = 8g × d, so d = 4g/8g = 0.5 m. The answer 1.5 m is the distance from B: the centre of mass lies nearer the end with the larger tension.

A block of mass 6 kg is placed at rest on a rough plane inclined at 20° to the horizontal. The coefficient of friction is 0.5. Taking g = 9.8 m s⁻², find the friction force acting on the block.

• A27.6 N
• B20.1 N✓
• C29.4 N
• D55.3 N

Answer: Maximum friction is μR = 0.5 × 6g cos 20° = 27.6 N. The component of weight down the slope is 6g sin 20° = 20.1 N, which is less than 27.6 N, so the block stays at rest and friction only needs to be 20.1 N. Quoting 27.6 N assumes friction is limiting when the block is not on the point of slipping.

A box of mass 4 kg is pulled up a rough slope inclined at 25° to the horizontal by a force of 50 N parallel to the slope. The coefficient of friction is 0.3. Taking g = 9.8 m s⁻², find the acceleration of the box, to 3 significant figures.

• A5.69 m s⁻²✓
• B8.36 m s⁻²
• C5.42 m s⁻²
• D2.38 m s⁻²

Answer: R = 4g cos 25° = 35.53 N, so friction = 0.3 × 35.53 = 10.66 N down the slope. Resolving up the slope: 50 − 4g sin 25° − 10.66 = 4a, so 50 − 16.57 − 10.66 = 22.78 = 4a and a = 5.69 m s⁻². Taking the normal reaction as 4g (not 4g cos 25°) gives 5.42 m s⁻².

A particle of mass 5 kg is on a rough plane inclined at 35° to the horizontal, with coefficient of friction 0.2. A horizontal force of 60 N, in the vertical plane containing a line of greatest slope, pushes the particle up the plane. Taking g = 9.8 m s⁻², find its acceleration, to 3 significant figures.

• A2.60 m s⁻²
• B3.98 m s⁻²
• C1.23 m s⁻²✓
• D4.77 m s⁻²

Answer: Perpendicular to the plane: R = 5g cos 35° + 60 sin 35° = 40.14 + 34.41 = 74.55 N, so friction = 0.2 × 74.55 = 14.91 N. Along the plane: 60 cos 35° − 5g sin 35° − 14.91 = 5a, i.e. 49.15 − 28.11 − 14.91 = 6.13, so a = 1.23 m s⁻². Leaving the 60 sin 35° term out of R gives 2.60 m s⁻²; the horizontal push presses the particle into the plane.

A particle is projected up a line of greatest slope of a rough plane inclined at 15° to the horizontal, with initial speed 8 m s⁻¹. The coefficient of friction is 0.4. Taking g = 9.8 m s⁻², find how far the particle travels before coming to rest, and state what happens next.

• A5.06 m; it then slides back down the plane
• B5.06 m; it then remains at rest on the plane✓
• C12.6 m; it then remains at rest on the plane
• D8.45 m; it then slides back down the plane

Answer: Moving up, friction and the weight component both act down the slope: deceleration = g(sin 15° + 0.4 cos 15°) = 6.323 m s⁻². Using v² = u² + 2as, 0 = 64 − 2(6.323)s, so s = 5.06 m. At rest, the weight component mg sin 15° is less than the maximum friction 0.4mg cos 15°, because tan 15° = 0.268 < 0.4, so it stays put. 12.6 m ignores friction.