High School

A 31.0 cm long spring is hung vertically from a ceiling and stretches to 37.9 cm when an 8.00 kg mass is hung from its free end.

(a) Find the spring constant (in N/m).
Spring constant: __________ N/m

(b) Find the length of the spring (in cm) if the 8.00 kg weight is replaced with a 205 N weight.
Length of the spring: __________ cm

Answer :

The spring constant is 115.94 N/m.

The length of the spring is 176.8 cm when the 8.00 kg weight is replaced with a 205 N weight.

(a) The spring constant of the given spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement. The equation for Hooke's Law is F = kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.

In this case, the displacement of the spring is given as 37.9 cm - 31.0 cm = 6.9 cm = 0.069 m. The force applied to the spring can be calculated using Newton's second law: F = mg, where m is the mass (8.00 kg) and g is the acceleration due to gravity (9.8 m/s²).

Substituting the values into Hooke's Law, we have 8.00 kg × 9.8 m/s² = k × 0.069 m.

(b) To find the length of the spring when a 205 N weight is hung on it, we can use Hooke's Law again. The force applied to the spring is now 205 N. We can rearrange Hooke's Law to solve for the displacement: x = F / k.

Substituting the values, we have x = 205 N / 115.94 N/m = 1.768 m.

Know more about Newton's second law here;

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