These eight original questions follow the lessons in force and motion II. Draw the diagram first, then write your answer, then open the explanation. Use g = 10 m s⁻².
If you need a lesson, start from the force and motion II hub and choose the skill that matches your mistakes.
Questions
Question 1. A 40 N force acts at 60° above the horizontal. Find its horizontal and vertical components. Use cos 60° = 0.50 and sin 60° = 0.87.
Answer
Horizontal = 40 cos 60° = 20 N. Vertical = 40 sin 60° = 40 × 0.87 = 35 N (34.8 N to three significant figures). See resolving forces into components.
Question 2. A 10 kg block rests on a slope of 37°. Find the weight component along the slope. Use sin 37° = 0.60.
Answer
Weight = 100 N. Component along the slope = 100 sin 37° = 60 N. The component perpendicular to the slope is 100 cos 37° = 80 N.
Question 3. Two forces of 5.0 N and 12 N act at right angles. Find the resultant and its direction relative to the 12 N force.
Answer
Resultant = √(25 + 144) = √169 = 13 N. Direction: tan θ = 5.0 ÷ 12, so θ = 22.6° from the 12 N force towards the 5.0 N force.
Question 4. Two forces of 8.0 N act on a point with 120° between them. Find the resultant.
Answer
The diagonal of the parallelogram splits the 120° angle into two 60° angles. The resultant = 2 × 8.0 × cos 60° = 2 × 8.0 × 0.50 = 8.0 N. See combining forces and finding a resultant.
Question 5. A sign of weight 30 N hangs from two strings, each at 30° to the horizontal. Find the tension in each string.
Answer
Vertical balance: 2T sin 30° = 30, so T = 30 N. The half-weight rule (15 N) would be wrong because the strings are angled. See applying equilibrium conditions.
Question 6. A spring extends by 6.0 cm under a 9.0 N load. Find k and the energy stored.
Answer
x = 0.060 m. k = 9.0 ÷ 0.060 = 150 N m⁻¹. Energy = ½ × 9.0 × 0.060 = 0.27 J. See using Hooke’s law and elastic energy.
Question 7. A spring with k = 300 N m⁻¹ is compressed by 5.0 cm and launches a 0.15 kg ball. Find the launch speed, ignoring losses.
Answer
Energy = ½ × 300 × 0.050² = 0.375 J. Then ½ × 0.15 × v² = 0.375, so v² = 5.0 and v = 2.2 m s⁻¹.
Question 8. A spring is tested: 0 N, 2 N, 4 N, 6 N, 8 N give extensions 0, 2.0, 4.0, 6.0, 10.0 cm. Find k from the straight part and say where the limit of proportionality lies.
Answer
From 0 to 6 N, extension rises 1.0 cm per 1 N, so k = 6 ÷ 0.060 = 100 N m⁻¹. The limit lies at about 6 N. From 6 N to 8 N, the extension rises 4.0 cm for 2 N, which is twice the expected 2.0 cm. See interpreting force-extension graphs.
If you got these wrong
Match each question to a lesson: 1 and 2 to components, 3 and 4 to resultants, 5 to equilibrium, 6 and 7 to Hooke’s law, and 8 to the graph lesson. If the diagrams themselves are unclear, revise drawing force diagrams.
Record errors with the mistake log and paper-error review tool, and rehearse with the timed original practice session builder. If you want a teacher to work through your diagrams with you, see online one-to-one Physics tuition.