### Question Description

Two springs of negligible weight and of spring constants 100N/M respectively are connected end to end and suspended from a fixed point. Determine the total extension when a mass of 7.5 kg is hung from the lower end.

Show all your workings. Answer must be correct

## Final Answer

Attached.

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Calculations on Hooke’s Law;

𝐹 = 𝑘𝑒, where F is force acting on the spring, k is the spring constant, e is the extension.

𝐹 = 𝑚𝑔 = 75 × 10 = 75𝑁

Therefore,

75 = 100 × 𝑒

Therefore, e will be given by

75

𝑒 = 100 = 0.75𝑚

When spri...

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