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If the length of a #32 cm# spring increases to #71 cm# when a #4 kg# weight is hanging from it, what is the spring's constant?

Answer 1

Charles Aeschliman

#k~=100,62 N/m#

#Delta x=71-32=39 " cm"=0,39" m"# #G=m*g=4*9,81=39,24N# #G=k*Delta x# #39,24=k* 0,39# #k~=100,62 N/m#

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Answer 2

Elena Albarran

The spring constant (k) can be calculated using Hooke's Law formula: ( k = \frac{F}{x} ), where F is the force applied (weight) and x is the displacement (change in length of the spring). Given: Force (F) = Weight = 4 kg * 9.8 m/s² (acceleration due to gravity) = 39.2 N Displacement (x) = Change in length = Final length - Initial length = 71 cm - 32 cm = 39 cm = 0.39 m Substitute values into the formula: ( k = \frac{39.2 , N}{0.39 , m} = 100.51 , N/m ) Therefore, the spring constant is approximately 100.51 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.