Question

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.

Hint:

In problems related to the elongation of strings due to tension, the relationship between tension and elongation is often linear, and the constant of proportionality (spring constant) can be used to find the original length.

Answer 1

Correct Answer: 1

Provided:

• Tension \(T_1 = 5\,\text{N}\) corresponds to length \(L_1 = 1.40\,\text{m}\).
• Tension \(T_2 = 7\,\text{N}\) corresponds to length \(L_2 = 1.56\,\text{m}\).

Objective: Determine the original (unstretched) length \(L_0\) of the string.

Concept Utilized:

For an elastic string, the extension \(x = L - L_0\) is directly proportional to the applied tension \(T\):

\[ T = kx = k(L - L_0) \]

where \(k\) represents the force constant of the string.

Solution Procedure:

Step 1: Formulate two equations based on the given data:

\[ T_1 = k(L_1 - L_0), \quad T_2 = k(L_2 - L_0) \]

Step 2: Eliminate the force constant \(k\) by dividing the equations:

\[ \frac{T_1}{L_1 - L_0} = \frac{T_2}{L_2 - L_0} \]

Step 3: Substitute the provided numerical values:

\[ \frac{5}{1.40 - L_0} = \frac{7}{1.56 - L_0} \]

Step 4: Solve for \(L_0\) by cross-multiplication and algebraic manipulation:

\[ 5(1.56 - L_0) = 7(1.40 - L_0) \] \[ 7.8 - 5L_0 = 9.8 - 7L_0 \] \[ 7L_0 - 5L_0 = 9.8 - 7.8 \] \[ 2L_0 = 2.0 \] \[ L_0 = 1.0\,\text{m} \]

Final Result:

\[ \boxed{L_0 = 1.0\,\text{m}} \]