A particle of mass $2$ kg hangs at rest from a spring of natural length $1$ m and modulus of elasticity $\lambda$ , fixed at O. At rest the stretch is $0.1$ m. Find $\lambda$ in terms of $g$ .

Model answer

20g

Also accepted: 20 g, 20g N, lambda = 20g

Explanation

Hooke's law gives the tension as

T = \dfrac{\lambda e}{l}

, where

e

is the extension and

l

the natural length. At rest the tension balances the weight:

\dfrac{\lambda(0.1)}{1} = 2g

, so

\lambda = 20g

newtons.

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q4

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