Grade 11 Physical Sciences · Mechanics

Apply Newton's second law

Use F_net = ma to link the net force on an object to its mass and acceleration (a 2 kg box pushed with 10 N speeds up at 5 m/s²).

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Grade 11 Newton Laws: Fnet = maKevinmathscience

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Question 2.3 · May/June 2025 · Paper 1 · National (DBE) · 3 marks

A crate of mass

25

kg slides down a rough inclined plane at a constant speed of

2

m·s

^{- 1}

towards point A while force F acts on it, parallel to the incline, as shown below. A constant kinetic frictional force of

40

N acts on the crate when the plane is inclined at an angle of

30°

with the horizontal. Diagram for this question

2.3 Calculate the magnitude of F.(3)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: apply newton's second law

Work each one out first, then open the memo and check every tick.

3. Question 4.3.1 · November 2025 · Paper 1 · Provincial · 3 marks
A boy pulls a $10$ kg block, that is attached to a $5$ kg block by a light, inextensible string of negligible mass, with a force of $50$ N at an angle of $30°$ with the horizontal as shown in the diagram below. The coefficient of kinetic friction between the surface and the $10$ kg block is $0, 18$ . The $5$ kg block experiences a kinetic frictional force of $9$ N.
4.3 Calculate the magnitude of the:
4.3.1 Normal force exerted on the $10$ kg block(3)
Show the memo (3 ticks)Memo
1. ✓
  Expression for N $N = F_{g} - F sin θ$
  1 mark · Apply Newton's second law
2. ✓
  Substitution $N = 10 \times 9, 8 - 50 sin 30°$
  1 mark · Apply Newton's second law
3. ✓
  Answer with unit $N = 73$ N
  1 mark · Apply Newton's second law
4. Question 5.4 · November 2022 · Paper 1 · National (DBE) · 5 marks
A $12$ kg block is initially at rest at point A at the bottom of a ROUGH inclined plane. The block is pulled up the incline by a constant force F acting parallel to the incline. The block reaches point B, which is at a vertical height of $4, 5$ m above the horizontal, with a speed of $2, 25$ m·s $^{- 1}$ . See the diagram below.
The same constant force F now moves the block at a CONSTANT VELOCITY across a rough horizontal surface from point B to point C, as shown below. Force F acts parallel to the horizontal surface.

The magnitude of the constant frictional force acting on the block while moving from point B to point C is $42$ N LARGER than the magnitude of the constant frictional force acting on the block when it moves from point A to point B.
5.4 Calculate the distance from point A to point B.(5)
Show the memo (5 ticks)Memo
1. ✓
  Formula for $W_{n c}$ Along the incline (AB):
  $W_{n c} = W_{F} + W_{f}$ (any one: $W_{n c} = F Δ x cos 0° + f_{1} Δ x cos 180°$ )
  1 mark · Calculate work done by a force
2. ✓
  Substitution of $559, 58$ J $559, 58 = (F - f_{1}) Δ x$ ... (1)
  1 mark · Calculate work done by a force
3. ✓
  Force equation on the horizontal (constant velocity)Along the horizontal (BC):
  $F - f_{2} = ma$
  $F - f_{2} = 0$ (OR $F = f_{2}$ )
  1 mark · Apply Newton's second law
4. ✓
  Relating the two frictional forces $F - (f_{1} + 42) = 0$
  $F - f_{1} = 42$ ... (2)
  1 mark · Apply Newton's second law
5. ✓
  AnswerSubstitute (2) into (1):
  $559, 58 = 42Δ x$
  $Δ x = 13, 32$ m
  1 mark · Calculate work done by a force
5. Question 7.4 · November 2022 · Paper 1 · National (DBE) · 5 marks
A charged sphere M is suspended from a ceiling by a light inextensible, insulated string.

Another charged sphere N, of mass $2, 04 \times 1 0^{- 3}$ kg and carrying a charge of $+ 8, 6 \times 1 0^{- 8}$ C, hangs STATIONARY vertically below sphere M. The centres of the spheres are $0, 3$ m apart, as shown in the diagram below.
7.4 Calculate the magnitude of the charge on sphere M.(5)
Show the memo (5 ticks)Memo
1. ✓
  Weight of N $F_{g} = m g = (2, 04 \times 1 0^{- 3}) (9, 8) = 0, 02$ N
  1 mark · Calculate weight
2. ✓
  Coulomb's formula $F = \frac{k Q _{M} Q _{N}}{r ^{2}}$
  1 mark · Apply Coulomb's law
3. ✓
  Equating forces ( $F_{n e t} = 0$ ) $F_{n e t} = m g - \frac{k Q _{M} Q _{N}}{r ^{2}}$
  $0 = 0, 02 - \frac{k Q _{M} Q _{N}}{r ^{2}}$ (OR $m g = \frac{k Q _{M} Q _{N}}{r ^{2}}$ )
  1 mark · Apply Newton's second law
4. ✓
  Substitution into Coulomb's law $\frac{( 9 \times 1 0 ^{9} ) Q _{M} ( 8 , 6 \times 1 0 ^{- 8} )}{( 0 , 3 ) ^{2}}$
  1 mark · Apply Coulomb's law
5. ✓
  Answer $Q_{M} = 2, 33 \times 1 0^{- 6}$ C
  1 mark · Apply Coulomb's law
6. Question 4.2.2 · May/June 2023 · Paper 1 · National (DBE) · 3 marks
Trolley A of mass $7, 2$ kg moves to the right at $0, 4$ m·s $^{- 1}$ in a straight line on a horizontal floor. It collides with a stationary trolley B of mass $5, 3$ kg.

After the collision, the trolleys lock together and move to the right, as shown in the diagram below.

Ignore any frictional effects.
4.2 Calculate the magnitude of the:
4.2.2 Average net force exerted by trolley A on trolley B during the collision, if the collision time is $0, 02$ s(3)
Show the memo (3 ticks)Memo
1. ✓
  METHOD 1: FormulaForce of A on B, right as positive:
  $F_{n e t} Δ t = Δ p$
  $F_{n e t} Δ t = m (v_{f} - v_{i})$ (any one)
  1 mark · Use the impulse-momentum theorem
2. ✓
  METHOD 1: Substitution $F_{n e t} (0, 02) = 5, 3 (0, 23 - 0)$
  1 mark · Use the impulse-momentum theorem
3. ✓
  METHOD 1: Answer with unit $F_{n e t} = 60, 95$ N
  1 mark · Use the impulse-momentum theorem

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