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If a spring has a constant of #12 (kg)/s^2#, how much work will it take to extend the spring by #77 cm #?

Answer 1

Charlotte Aaron

The work done is #=3.56J#

The effort is equivalent to the spring's stored energy.

The spring constant is #k=12kgs^-2#

The extension is #x=0.77m#

The energy that has been stored is

#W=1/2kx^2=1/2*12*0.77^2=3.56J#

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Answer 2

Adam Arnold

The work done to extend the spring is given by ( W = \frac{1}{2}kx^2 ), where ( k ) is the spring constant and ( x ) is the displacement from the equilibrium position. Calculate the work using this formula.

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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.