Question

question 20
1 pts
a spring is hung from the ceiling and a small box is attached to the bottom of the spring causing it to stretch a distance of 0.47 meters. the spring has a spring constant k = 6.9 n/m. assuming the box is at rest, calculate the mass of the box needed to cause the spring to stretch to this distance.
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Explanation:

Step1: Determine the force exerted by the spring

According to Hooke's law, \( F = kx \), where \( k = 6.9\,\text{N/m} \) and \( x = 0.47\,\text{m} \).

$$ F=6.9\times0.47 = 3.243\,\text{N} $$

Step2: Relate the force to the weight of the box

When the box is at rest, the force exerted by the spring \( F \) is equal to the weight of the box \( W = mg \), where \( g = 9.8\,\text{m/s}^2 \).

$$ m=\frac{F}{g} $$

Substitute \( F = 3.243\,\text{N} \) and \( g = 9.8\,\text{m/s}^2 \) into the formula:

$$ m=\frac{3.243}{9.8}\approx0.33\,\text{kg} $$

Answer:

\( 0.33\,\text{kg} \)