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Answer :
The potential energy of the triceps muscle when using the rubber band with a spring constant [tex]\( k = 25.0 \, \text{N/m} \)[/tex] and stretched to a displacement [tex]\( x = 35.9 \, \text{mm} \)[/tex]is approximately[tex]\( 2.06495 \times 10^{-5} \, \text{N} \cdot \text{m} \).[/tex]
To find the potential energy stored in the triceps muscle when using a rubber band with a spring constant [tex]\( k = 25.0 \, \text{N/m} \)[/tex] and stretched to a displacement [tex]\( x = 35.9 \, \text{mm} \),[/tex] we can use the formula for the potential energy stored in a spring:
[tex]\[ U = \frac{1}{2} k x^2 \][/tex]
where:
- [tex]\( U \)[/tex] is the potential energy stored in the spring,
- [tex]\( k \)[/tex] is the spring constant,
- [tex]\( x \)[/tex] is the displacement from the equilibrium position.
First, let's convert the displacement [tex]\( x \)[/tex] from millimeters to meters:
[tex]\[ x = 35.9 \, \text{mm} = 35.9 \times 10^{-3} \, \text{m} \][/tex]
Now, we can use the formula to find the potential energy:
[tex]\[ U = \frac{1}{2} \times 25.0 \, \text{N/m} \times (35.9 \times 10^{-3} \, \text{m})^2 \]\[ U = \frac{1}{2} \times 25.0 \times (1.28721 \times 10^{-3})^2 \, \text{N} \cdot \text{m} \]\[ U = \frac{1}{2} \times 25.0 \times 1.65196 \times 10^{-6} \, \text{N} \cdot \text{m} \]\[ U = \frac{1}{2} \times 4.1299 \times 10^{-5} \, \text{N} \cdot \text{m} \]\[ U = 2.06495 \times 10^{-5} \, \text{N} \cdot \text{m} \][/tex]
So, the potential energy of the triceps muscle when using the rubber band with a spring constant [tex]\( k = 25.0 \, \text{N/m} \)[/tex] and stretched to a displacement [tex]\( x = 35.9 \, \text{mm} \)[/tex]is approximately[tex]\( 2.06495 \times 10^{-5} \, \text{N} \cdot \text{m} \).[/tex]
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