If the length of a #47# #cm# spring increases to #89# #cm# when a #2# #kg# weight is hanging from it, what is the spring's constant?

Answer 1

We should work in SI units, so the stretch is #0.89-0.47=0.42# #m#. Then we know the weight force of a #2# #kg# mass is #19.6# #N# and #F=kx#, and can rearrange to give: #k=F/x=19.6/0.42=46.7# #Nm^-1 or kgs^-2#.

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Answer 2

To find the spring constant (k), you can use Hooke's Law: F = kx, where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.

First, calculate the force applied using the weight and gravity: F = mg, where m is the mass (2 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Then, calculate the displacement (x) of the spring: x = (final length - initial length) = (89 cm - 47 cm).

Once you have the force and displacement, you can solve for the spring constant (k) using Hooke's Law: k = F / x.

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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.

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.

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.

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.

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