A particle of mass $2$ kg hangs from a spring of natural length $1$ m and modulus $20g$ , fixed at O, and rests with a stretch of $0.1$ m. It is then pulled down a further $x$ metres. Which expression gives the tension in the spring in that pulled-down position?

2g + 20gx

, since the total extension is now

0.1 + x

and

T = \dfrac{\lambda(0.1+x)}{1}

20gx

, since only the extra stretch

x

beyond the resting position contributes to the tension.

\dfrac{20g}{0.1 + x}

, since the tension falls as the total extension of the spring grows.

2g - 20gx

, since the weight of the particle acts against the pull of the stretched spring.

Explanation

Hooke's law uses the total extension measured from the natural length, not the extra displacement. Here the total extension is

0.1 + x

, so

T = \dfrac{20g(0.1 + x)}{1} = 2g + 20gx

. The constant part

2g

is exactly the weight, which is why it cancels when the equation of motion is written, leaving a restoring force proportional to

x

alone.

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

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