To find the work done by the spring while it is being extended further by an external force, we can use the concept of spring potential energy and the work done by a spring.
The work done by a spring when it is stretched or compressed by a distance \( x \) from its equilibrium position is given by the formula:
\[ W = -\frac{1}{2} k x^2 \]
where:
- \( W \) is the work done by the spring,
- \( k \) is the spring constant, and
- \( x \) is the displacement from the spring's natural length.
Step 1: Determine the spring constant \( k \)
We know that when a 2.0 kg mass is hung from the spring, it extends 6.0 cm due to the weight of the mass. The weight of the mass can be calculated as:
\[ F = m \cdot g = 2.0 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 19.62 , \text{N} \]
At equilibrium, the force exerted by the weight of the mass equals the spring force:
\[ F_s = k \cdot x \]
Substituting in the known values:
\[ 19.62 , \text{N} = k \cdot 0.06 , \text{m} \]
Solving for \( k \):
\[ k = \frac{19.62 , \text{N}}{0.06 , \text{m}} \approx 327 , \text{N/m} \]
Step 2: Calculate the work done by the spring for the additional extension
Now, when an additional 10 cm (0.10 m) is applied downward, the total displacement from the spring's relaxed length becomes:
\[ x_{\text{total}} = 6.0 , \text{cm} + 10.0 , \text{cm} = 16.0 , \text{cm} = 0.16 , \text{m} \]
The work done by the spring as it is extended from 6.0 cm to 16.0 cm (the additional 10.0 cm stretch) can be calculated using the spring's work formula for the range of extension from \( x_1 = 0.06 , \text{m} \) to \( x_2 = 0.16 , \text{m} \):
\[ W = -\frac{1}{2} k x_2^2 + \frac{1}{2} k x_1^2 \]
Calculating each term:
\[ W = -\frac{1}{2} \cdot 327 , \text{N/m} \cdot (0.16)^2 + \frac{1}{2} \cdot 327 , \text{N/m} \cdot (0.06)^2 \]
Calculating \( (0.16)^2 \):
\[ (0.16)^2 = 0.0256 , \text{m}^2 \]
Calculating \( (0.06)^2 \):
\[ (0.06)^2 = 0.0036 , \text{m}^2 \]
Now substituting these values back into the equation for work:
\[ W = -\frac{1}{2} \cdot 327 \cdot 0.0256 + \frac{1}{2} \cdot 327 \cdot 0.0036 \]
Calculating each term:
\[ W = -\frac{1}{2} \cdot 327 \cdot 0.0256 = -4.18 , \text{J} \quad \text{(approximately)} \]
\[ W = \frac{1}{2} \cdot 327 \cdot 0.0036 = 0.59 , \text{J} \quad \text{(approximately)} \]
Thus:
\[ W \approx -4.18 , \text{J} + 0.59 , \text{J} = -3.59 , \text{J} \]
Therefore, the work done by the spring while it is being extended an additional 10 cm is approximately:
\[ \boxed{-3.59 , \text{J}} \]
This indicates that the work done by the spring is negative, which is consistent with the spring acting to resist being stretched.
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