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If the length of a #46 cm# spring increases to #57 cm# when a #8 kg# weight is hanging from it, what is the spring's constant?

Answer 1

Adrian Ayala

#"713 N/m"#

#F = kDeltax#

#k = F/(Deltax) = (mg)/(Deltax) = ("8 kg × 9.8 m/s"^2)/("0.57 m - 0.46 m") = "713 N/m"#

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Answer 2

Axel Acevedo

To find the spring constant (k), we can use Hooke's Law equation:

F = k * x

Where: F is the force applied to the spring (weight hanging from it) k is the spring constant x is the displacement from the equilibrium position

Given: Force (F) = weight = mass * acceleration due to gravity = 8 kg * 9.8 m/s^2 = 78.4 N Initial length of the spring (x_initial) = 46 cm = 0.46 m Final length of the spring (x_final) = 57 cm = 0.57 m

Change in length (Δx) = x_final - x_initial Δx = 0.57 m - 0.46 m = 0.11 m

Using Hooke's Law:

78.4 N = k * 0.11 m

Solve for k:

k = 78.4 N / 0.11 m = 712.73 N/m

So, the spring constant is approximately 712.73 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.