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

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:

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