IGCSE Physics 1.5 — Forces Numericals (Solutions)

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IGCSE Physics 0625 · Section 1.5 · Forces — Numerical Solutions

Forces — Numerical Practice: Solutions

A. Resultant Force and F = ma

A1

Resultant 20 N on 4 kg. Find acceleration.

1
a = F/m
20 ÷ 4
✓5 m/s²
A2

3 kg, a = 6 m/s². Find resultant force.

1
F = ma
3 × 6
✓18 N
A3

50 N forward, 20 N backward. Find resultant.

1
Subtract opposing forces
50 − 20
✓30 N forward
A4

1000 kg, resultant force 2500 N. Find acceleration.

1
a = F/m
2500 ÷ 1000
✓2.5 m/s²
A5

8 kg, a = 3 m/s² needed, driving force 40 N. Find resistive force.

1
Resultant force needed
F = ma = 8 × 3 = 24 N
2
Resistive force = driving force − resultant needed
40 − 24
✓16 N
A6

15 N and 9 N opposite directions. Find resultant.

1
Subtract the smaller from the larger
15 − 9
✓6 N, in the direction of the 15 N force
A7

2000 kg, 0 to 15 m/s in 5 s. Find resultant force.

1
Find acceleration first
a = 15 ÷ 5 = 3 m/s²
2
F = ma
2000 × 3
✓6000 N

B. Spring Constant

B1

Extends 4 cm under 8 N. Find k.

1
k = F/x
8 ÷ 4
✓2 N/cm
B2

k = 3 N/cm, load 15 N. Find extension.

1
x = F/k
15 ÷ 3
✓5 cm
B3

k = 4 N/cm, extension 7 cm. Find force.

1
F = kx
4 × 7
✓28 N
B4

k = 2.5 N/cm, extension 6 cm. Find force.

1
F = kx
2.5 × 6
✓15 N
B5

Straight line to 10 N / 5 cm, then curves. Find k, explain the curve.

1
k = F/x in the straight-line region
10 ÷ 5
✓k = 2 N/cm. Beyond 10 N, the spring has passed its limit of proportionality — extension is no longer proportional to load.

C. Moments

C1

20 N at 0.5 m from pivot. Find moment.

1
moment = F × d
20 × 0.5
✓10 N m
C2

Moment 15 N m at 0.3 m. Find force.

1
F = moment/d
15 ÷ 0.3
✓50 N
C3

300 N at 2 m balances see-saw; find weight at 3 m.

1
Clockwise moment = anticlockwise moment
300 × 2 = W × 3
2
Solve for W
600 ÷ 3
✓200 N
C4

20 N at 0.4 m (left); find force at 0.8 m (right) to balance.

1
Set moments equal
20 × 0.4 = F × 0.8
2
Solve for F
8 ÷ 0.8
✓10 N
C5

Left: 10 N at 0.5 m and 5 N at 1 m. Find force at 0.6 m (right) to balance.

1
Total anticlockwise moment (left)
(10 × 0.5) + (5 × 1) = 5 + 5 = 10 N m
2
Set equal to the clockwise moment and solve for F
F × 0.6 = 10 → F = 10 ÷ 0.6
✓≈ 16.7 N
C6

Two conditions for equilibrium.

✓No resultant force acting on the object, AND no resultant moment acting on the object.
C7

15 N at 0.8 m; find moment, then force needed at 0.4 m for the same moment.

1
moment = F × d
15 × 0.8 = 12 N m
2
Rearrange for the new force at 0.4 m
F = 12 ÷ 0.4
✓Moment = 12 N m; force needed at 0.4 m = 30 N
C8

40 N at 0.15 m. Find moment.

1
moment = F × d
40 × 0.15
✓6 N m

Mastery Test — Answer Key

  1. 6 m/s² (30 ÷ 5)
  2. 15 N (6 × 2.5)
  3. 30 N — in the direction of the 45 N force (45 − 15)
  4. 4 cm (20 ÷ 5)
  5. 3 N/cm (18 ÷ 6)
  6. 10 N m (25 × 0.4)
  7. 150 N — 250×1.5 = W×2.5 → W = 375÷2.5
  8. 20 N — 12×0.5 = F×0.3 → F = 6÷0.3
  9. No resultant force acting on the object, AND no resultant moment acting on the object.
  10. 12 N m (20 × 0.6)

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