Question

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a spring is initially compressed 2.08 cm and fires a 12.3-gram dart straight up into the air. If the dart reaches a height of 3.25 m, determine spring constant.

Answer 1

The spring constant is 181.0 N/m

Explanation:

We can solve the problem by applying the law of conservation of energy. In fact, the elastic potential energy initially stored in the compressed spring is completely converted into gravitational potential energy of the dart when the dart is at its maximum height. Therefore, we can write:

$\frac{1}{2}kx^2 = mgh$

where the term on the left represents the elastic potential energy of the spring while the term on the right is the gravitational potential energy of the dart at maximum height, and where

k is the spring constant of the spring

x = 2.08 cm = 0.0208 m is the compression of the spring

m = 12.3 g = 0.00123 kg is the mass of the dart

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 3.25 m is the maximum height of the dart

Solving for k, we find:

$k=\frac{2mgh}{x^2}=\frac{2(0.00123)(9.8)(3.25)}{(0.0208)^2}=181.0 N/m$

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