If the length of a #13 cm# spring increases to #40 cm# when a #7 kg# weight is hanging from it, what is the spring's constant?

Answer 1

I got: #k=254N/m#

Consider that at equilibrium the elastic force #F_(el)# (given by Hooke's Law: #F_(el)=-kx#) and the weight #W# must balance to give, into Newton's Second Law, zero acceleration (the sistem is at rest). We get: #F_(el)+W=ma=0# #-kx-mg=0# (eleasic force upwards and weight downwards) The displacement #x# is considered negative (downwards) so we have: #-k[-(0.4-0.13)]-(7*9.8)=0# (I changed into meters) rearranging we get: #k=254N/m#

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Answer 2

Sebastian Acosta

The spring constant is approximately 245 N/m.

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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.