1. State Hooke’s law (1mk)
✓For a helical spring or any other elastic material, extension is directly
proportional to the stretching force,
... [Show More] provided elastic limit is not
exceeded
2. Define the term elastic limit, as used in stretching of materials (1mk)
✓ The point beyond which the elastic material does not obey Hooke’s law
3. State the SI units of elastic constant of a spring (1mk)
✓ Newton per meter (N/m)
4. State the features that govern the strength of a spiral spring of a given
material. (2mks)
Strength of a spring is the ability of a material to resist breakage when under
stretching, compressing or shearing force. A strong material is one which can
withstand a large force without breaking. Hence the strength depends on:
i. Diameter of the spring
ii. Diameter of the wire making the spring
iii. Number of the coils of the spring
iv. Nature of the material making the spring
5. State two factors on which the extension of a wire depends on assuming it
obeys Hooke’s Law. (2mks)
i. Length of the wire
ii. Load
iii. Nature of the wire (spring constant)
6. Apart from the diameter and length, name another factor that determines the
spring constant of a spiral spring. (1mk)
i. Type of material making the wire
ii. The number of turns per unit length of the spring
7. A heavy load is suspended on a wire. Give any one factor that will determine
extension in the wire. (1mk)
i. Length of the wire
ii. Nature of the wire (spring constant)
8. Distinguish between ductile end and brittle material
Ductile materials elongate considerably when under stretching forces and
undergo plastic deformation until they break e.g. lead, copper, while Brittle
materials do not undergo extension and break without warning on
stretching. E.g. blackboard chalk, bricks, cast iron, glass, and dry biscuits.
9. It is easier to bend an iron rod than a glass rod of the same dimensions at
room temperature. Give a reason for this
Iron is a ductile material while glass is a brittle material.
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10. A spiral spring stretches by 0.6 cm when a mass of 300g is suspended on it.
What is the spring constant?
F=ke
e = 0.006m
F=mg= 0.3×10=3
𝑘 =
3
0.006
= 500N/m
11. An unloaded spring has a length of 15cm and when under a load of 24N it
has a length of 12cm.What will be the load on the spring when length is
10cm?
𝐹 = 𝑘𝑒
F=24N
E= 0.03m
k=
24
0.03
= 800N/m
F= 800×0.05
=40N
12. An object of weight 20N attached at the end of a spring causes an extension
of 0.5 cm on the spring.
(a) Determine the spring constant. (2mk)
K= 20
0.005
= 4000N/m
(b) Determine the weight of an object that would cause an extension of 0.86
cm on the same spring. (1mk)
𝐹 = 𝑘𝑒
= 4000×0.0086
= 3.44N
13. A copper wire is 2m long. A force of 4N suspended on the wire while other
end is fixed increases its length to 2.001m. What force would make the
length of the wire 2.032m? (3mks)
𝑘 =
𝐹
𝑒
=
4
0.001
= 4000N/m
𝐹 = 𝑘𝑒
= 4000×0.032
=128N
14. The pointer of an unloaded spring reads 32 cm. when a mass of 120g is
applied to the spring, the pointer reads 38cm. a pan in which a mass of
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210g is placed is now hang from the spring and the pointer reads 48cm.
determine the mass of the pan. (4mks)
F=1.2N and e= 0.06 hence 𝑘 =
𝐹
𝑒
=
1.2
0.06
= 20N/m
𝐹 = 𝑘𝑒
= 20×0.16
= 3.2N
Weight of the pan = 3.2-2.1 =1.1N
Mass of the pan = 1.1
10
= 0.11Kg = 110g
15. Two identical helical springs are connected in series. When a 50g mass is
hang at the end of the springs, it produces an extension of 2.5 cm.
Determine the extension produced by the same mass when the springs arc
connected in parallel. (3mks)
Ks =
𝐹
𝑒
where e is total extension
=
0.5
0.25
= 2 N/m
K1 = 4N/m
kp =8N/m
e =
𝐹
𝑘
=
0.5
8
= 0.0625m =6.25cm
16. Two springs of negligible weights and spring constants 50N/m and 75N/m
respectively are connected in series and suspended from a fixed point.
Determine the total extension when a mass of 7.5kg is hung from the lower
end. (3mk)
et = e1+e2
e1=
𝐹
K1
=
75
50
+ 75
75
= 1.5 +1
= 2.5m
17. Two springs of negligible weight and of spring constants 100Nm-1respectively
are connected end to end and suspended from a fixed point. Determine;
(i) The total extension when a mass of 7.5 kg is hung from the lower
end. (2mk)
et =
𝐹
𝑘𝑠
=
75
50
= 1.5m
(ii) The elastic constant of the combined of springs [Show Less]