Hooke's law holds extension proportional to force only within the elastic limit, and extension means the increase in length when a force is applied, so this Year 9 sheet opens with the law's limits and its measure. The key records F = k × e and N/m for the constant.
Key Takeaways
Extension is proportional to force within the elastic limit (Q1).
Extension is the increase in length when force is applied (Q2).
Q5 records F = k × e as the law.
Section A: Multiple Choice Questions (1 Mark Each)
Choose the correct option for each question.
1.Hooke's law states that the extension of a spring is proportional to the force applied
(a) within the elastic limit(b) for any force(c) only when the spring is compressed(d) only for very large forces
2.The extension of a spring is
(a) the increase in length when a force is applied(b) the total length of the spring(c) half the length of the spring(d) the mass of the spring
3.The unit of the spring constant k is
(a) N/m(b) N(c) m(d) N × m
4.Beyond the elastic limit, a spring
(a) is permanently deformed and no longer obeys Hooke's law(b) returns to its original length(c) becomes stronger(d) stops stretching completely
5.The formula for Hooke's law is
(a) F = k × e(b) F = k ÷ e(c) F = e ÷ k(d) F = m × g
6.Two identical springs arranged side by side (in parallel) and supporting the same load are
(a) stiffer, so they extend less(b) softer, so they extend more(c) equally stiff(d) unable to support any load
Section B: Short Answer Type Questions (2 Marks Each)
Show all steps clearly.
7.State Hooke's law.
8.Define extension.
9.Write the formula for Hooke's law and name each symbol.
10.What is the spring constant?
11.What happens when a spring is stretched beyond its elastic limit?
12.What is meant by the elastic limit?
13.Give one practical use of springs.
14.How is the spring constant k measured experimentally?
Section C: Numericals & Word Problems (3 Marks Each)
Apply the concepts to solve the problems. Show all working.
15.A spring extends by 4 cm when a force of 8 N is applied. Using k = F ÷ e, find the spring constant in N/cm.
16.A spring with a spring constant of 100 N/m is stretched by 0.2 m. Using F = k × e, find the force applied.
17.A force of 30 N produces an extension of 6 cm. Using k = F ÷ e, find the spring constant in N/cm.
18.A spring of natural length 20 cm extends to 26 cm when a force of 12 N is applied. Find the extension and then the spring constant.
19.A spring with k = 50 N/m extends by 0.4 m. Using F = k × e, find the force applied.
20.A spring extends by 3 cm under a force of 6 N. Using e = F ÷ k (or proportion), find the extension when the force is doubled to 12 N, within the elastic limit.
Answer Key
1. a) within the elastic limit
2. a) the increase in length when a force is applied
3. a) N/m
4. a) is permanently deformed and no longer obeys Hooke's law
5. a) F = k × e
6. a) stiffer, so they extend less
7. Refer to solution guide.
8. Refer to solution guide.
9. Refer to solution guide.
10. Refer to solution guide.
11. Refer to solution guide.
12. Refer to solution guide.
13. Refer to solution guide.
14. Refer to solution guide.
15. Refer to solution guide.
16. Refer to solution guide.
17. Refer to solution guide.
18. Refer to solution guide.
19. Refer to solution guide.
20. Refer to solution guide.
End of Worksheet
People Also Ask
When does Hooke's law apply?
Within the elastic limit, option (a) in this worksheet (Q1).
What is meant by extension?
The increase in length when a force is applied, option (a) in this worksheet (Q2).
How is Hooke's law written?
F = k × e, option (a) in this worksheet (Q5).
Stuck on a similar problem? Clear your next doubt instantly with the Doubt Solver.
Browse more free worksheets for IGCSE Year 9.