Grade 11 Physical Sciences · Mechanics

Calculate weight

Calculate weight from mass using w = mg (a 50 kg learner weighs 490 N), and tell mass in kg apart from weight in N.

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Grade 11 and 12 Forces: Weight and components of WeightMiss Martins Maths and Science

Physics - What Is The Difference Between Mass and Weight?The Organic Chemistry Tutor

Mass vs. Weight and the Normal ForceProfessor Dave Explains

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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. Diagram for this question

7.4 Calculate the magnitude of the charge on sphere M.(5)

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

More past paper questions: calculate weight

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3. Question 2.2.2 · November 2025 · Paper 1 · Provincial · 2 marks
Three forces, $F_{1}$ , $F_{2}$ and $F_{3}$ , NOT DRAWN TO SCALE, are applied to a crate of $25$ kg as shown in the diagram below.
2.2 Calculate the:
2.2.2 Magnitude of the gravitational force acting on the crate(2)
Show the memo (2 ticks)Memo
1. ✓
  Substitution $W = m g$
  $W = 25 \times 9, 8$
  1 mark · Calculate weight
2. ✓
  Answer with unit $W = 245$ N
  1 mark · Calculate weight
4. Question 7.2.3 · May/June 2025 · Paper 1 · National (DBE) · 5 marks
7.2 Sphere S is attached to the ceiling at point P by a light inextensible string. When sphere R, on an insulated stand, is brought close to sphere S they repel and sphere S comes to rest with the horizontal distance between the centres of the spheres equal to $0, 03$ m.

Sphere R is directly below point P. The string makes an angle θ with the vertical, as shown in the diagram below.

Answer the questions that follow.
Sphere R has a charge of $+ 6$ nC. Sphere S has a charge $+ 3$ nC and a mass of $0, 15$ g.
7.2.3 Calculate angle θ.(5)
Show the memo (5 ticks)Memo
1. ✓
  Formula $F = \frac{k Q _{1} Q _{2}}{r ^{2}}$
  1 mark · Apply Coulomb's law
2. ✓
  Substitution $F = \frac{( 9 \times 1 0 ^{9} ) ( 6 \times 1 0 ^{- 9} ) ( 3 \times 1 0 ^{- 9} )}{( 0 , 03 ) ^{2}}$
  $= 1, 8 \times 1 0^{- 4}$ N
  1 mark · Apply Coulomb's law
3. ✓
  Weight of S $w = m g = (1, 5 \times 1 0^{- 4}) (9, 8) = 1, 47 \times 1 0^{- 3}$ N
  1 mark · Calculate weight
4. ✓
  Trig ratio $tan θ = \frac{F _{E}}{m g} = \frac{1 , 8 \times 1 0 ^{- 4}}{1 , 47 \times 1 0 ^{- 3}}$
  1 mark · Find resultant of perpendicular vectors
5. ✓
  Answer $θ = 6, 98°$
  1 mark · Find resultant of perpendicular vectors
5. Question WE23 · Siyavula Grade 11 Physical Sciences, 2.4 Forces between masses, Worked example 23 · 3 marks
WE23 A man has a mass of $70$ kg. On the planet Beeble how much will he weigh if Beeble has a mass half of that of the Earth and a radius one quarter that of the Earth. Gravitational acceleration on Earth is $9, 8$ m·s $^{- 2}$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Weight on Earth $F_{E a r t h} = m g_{E a r t h} = (70) (9, 8)$
  $= 686$ N
  1 mark · Calculate weight
2. ✓
  Beeble in terms of Earth $F_{B ee b l e} = G \frac{M _{B ee b l e} m}{r _{B}^{2}} = G \frac{\frac{1}{2} M _{E a r t h} m}{( \frac{1}{4} r _{E} ) ^{2}}$
  $= 8 (G \frac{M _{E a r t h} m}{r _{E}^{2}})$
  1 mark · Apply Newton's law of gravitation
3. ✓
  Answer with unit $F_{B ee b l e} = 8 (686) = 5488$ N
  1 mark · Apply Newton's law of gravitation

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