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Expert: - 12/14/2011


Question
Suppose 2 Nm of work is needed to stretch a spring from its natural length of 0.3m to a length of .42m how much work is required to stretch the spring from 0.35m to .4m?

Answer
Potential energy stored in a spring when stretched by x m = (1/2)k x^2,
where k is the spring constant.

Hence, 2=(1/2)k (0.42-0.3)^2
      k=277.778 N/m

Amount of work needed to stretch spring by 0.4-0.35=0.05m
 =(1/2)*277.78*0.05^2= 0.3472 Nm (or Joules)    (shown)

Hope this helps. Peace.

(This is actually a physics question not a calculus -based one; perhaps posting it in the relevant section would enable you to get the assistance you require more efficiently in the future.)

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Frederick Koh

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I can answer questions concerning calculus, complex numbers, vectors, statistics , algebra and trigonometry for the O level, A level and 1st/2nd year college math/engineering student.

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3 years helping out in a singaporean youth forum: http://forums.sgclub.com/math_help/ (under the moniker whitecorp) You can also visit my main maths website http://www.whitegroupmaths.com where I have designed "question locker" vaults to store tons of fully worked math problems. Peace.

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