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A spring elongates from 13.00 cm to 14.50 cm when a 3.95 kg object is attached to it. What is the spring constant of the spring?

a) 96.0 N/m
b) 130.0 N/m
c) 25.0 N/m
d) 64.0 N/m

User Narxx
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1 Answer

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Final answer:

To find the spring constant, we can use Hooke's law which states that the force exerted by a spring is directly proportional to its displacement. Using the given information, we can calculate the spring constant to be approximately 2557.33 N/m.

Step-by-step explanation:

To find the spring constant, we can use Hooke's law which states that the force exerted by a spring is directly proportional to its displacement. Hooke's law can be written as F = -kx, where F is the force, k is the spring constant, and x is the displacement. We can use the given information to calculate the spring constant:

First, we need to convert the elongation of the spring into meters. The initial elongation is 13.00 cm = 0.13 m, and the final elongation is 14.50 cm = 0.145 m. The displacement (x) is the difference between the final and initial elongation, so x = 0.145 m - 0.13 m = 0.015 m.

Now, we can use Hooke's law to find the spring constant. Since the force is directly proportional to the displacement, we can write F = kx. Rearranging the equation, we have k = F / x. The force (F) can be calculated using the weight of the object (m * g, where m is the mass and g is the acceleration due to gravity).

Plugging in the values, we have: k = (m * g) / x = (3.95 kg * 9.8 m/s^2) / 0.015 m =~ 2557.33 N/m.

Therefore, the spring constant of the spring is approximately 2557.33 N/m.

User Jim Nutt
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