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JAMB Physics 2012 Paper

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Question : 10

Total: 42

If a load of 1 kg stretches a cord by 1.2cm, what is the force constant of the cord? {g=10ms^-2}

866Nm^-1
833Nm^-1
769Nm^-1
667Nm^-1

Solution:

Concept:

The force constant of a cord is found using Hooke's law:

F = k e

.

Explanation:

The load exerts a force equal to its weight.

So,

F = mg = 1 \times 10 = 10 \, \mathrm{N}

.

The extension is

e = 1.2 \, \mathrm{cm} = 0.012 \, \mathrm{m}

.

Using

k = \frac{F}{e}

,

k = \frac{10}{0.012}

.

k = \frac{10 \times 1000}{12}

.

k = 833.33 \, \mathrm{N\,m^{-1}}

This is approximately

833 \, \mathrm{N\,m^{-1}}

.

Answer:

B.

833 \, \mathrm{N\,m^{-1}}

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