The length of a light string is 1.4 m when the tension on it is 5 N. If the tension increases to 7 N, the length of the string is 1.56 m. The original length of the string is ______ m.
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Correct Answer: 1
Solution and Explanation
We are given:
- When tension \(T_1 = 5\,\text{N}\), length \(L_1 = 1.40\,\text{m}\).
- When tension \(T_2 = 7\,\text{N}\), length \(L_2 = 1.56\,\text{m}\).
We must find the original (unstretched) length \(L_0\) of the string.
Concept Used:
For an elastic string, extension \(x = L - L_0\) is proportional to the applied tension \(T\):
\[ T = kx = k(L - L_0) \]
where \(k\) is the force constant of the string.
Step-by-Step Solution:
Step 1: Write the two equations:
\[ T_1 = k(L_1 - L_0), \quad T_2 = k(L_2 - L_0) \]
Step 2: Divide or eliminate \(k\):
\[ \frac{T_1}{L_1 - L_0} = \frac{T_2}{L_2 - L_0} \]
Step 3: Substitute known values:
\[ \frac{5}{1.40 - L_0} = \frac{7}{1.56 - L_0} \]
Step 4: Cross multiply:
\[ 5(1.56 - L_0) = 7(1.40 - L_0) \] \[ 7L_0 - 5L_0 = 7(1.40) - 5(1.56) \] \[ 2L_0 = 9.8 - 7.8 = 2.0 \] \[ L_0 = 1.0\,\text{m} \]
Final Computation & Result:
\[ \boxed{L_0 = 1.0\,\text{m}} \]
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