If a spring has a constant of #12 (kg)/s^2#, how much work will it take to extend the spring by #76 cm #?

Answer 1

#4,56J#

I assume the units of the spring constant you meant was #12N//m# since from Hooke's Law, #F=kx=>k=F/x# and so the dimensions of k is N/m.

In this case, we use the formula for energy stored in a compressed or elongated spring. #W=1/2kx^2#, (this may be derived from calculus - ask me if you require the derivation as well), to obtain

#W=1/2xx12N//mxx0,76m#

#=4,56J#

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Answer 2

Emily Abbate

The work done to extend the spring by 76 cm is 4.56 joules.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.