Question

An elastic spring under tension of 3 N has a length a. Its length is b under tension 2 N. For its length (3a – 2b), the value of tension will be_____ N.

Answer 1

Correct Answer: 5

Given:

\[ 3 = K(a - \ell) \] \[ 2 = K(b - \ell) \]

where \( K \) is the spring constant and \( \ell \) is the natural length of the spring.

To find the tension \( T \) for the length \( (3a - 2b) \), we use:
\[ T = K (3a - 2b - \ell) \]

Substituting the values of \( a - \ell \) and \( b - \ell \) from the given equations:
\[ T = K [3(a - \ell) - 2(b - \ell)] \]

\[ T = K \left[ 3 \left( \frac{3}{K} \right) - 2 \left( \frac{2}{K} \right) \right] \]

\[ T = K \left[ \frac{9}{K} - \frac{4}{K} \right] \]

\[ T = K \left[ \frac{5}{K} \right] = 5 \, \text{N} \]

Conclusion:
Hence, the value of the tension is \( 5 \, \text{N} \).

Answer 2

Step 1: Given data.
Tension \( T_1 = 3 \, \text{N} \) corresponds to length \( a \).
Tension \( T_2 = 2 \, \text{N} \) corresponds to length \( b \).
We need to find the tension when the spring length is \( (3a - 2b) \).

Step 2: Apply Hooke’s Law.
For an elastic spring, extension is directly proportional to the applied force (tension):
\[ T = k(x - x_0) \] where \( x \) is the length under tension, and \( x_0 \) is the natural (unstretched) length.

Step 3: Write two equations.
\[ 3 = k(a - x_0) \quad \text{(1)} \] \[ 2 = k(b - x_0) \quad \text{(2)} \] Subtract (2) from (1):
\[ 3 - 2 = k(a - b) \] \[ k = \frac{1}{a - b} \]
Step 4: Find \( x_0 \) (natural length).
From equation (1):
\[ 3 = \frac{1}{a - b}(a - x_0) \] \[ a - x_0 = 3(a - b) \] \[ x_0 = a - 3(a - b) = a - 3a + 3b = 3b - 2a \]
Step 5: Find tension for length \( (3a - 2b) \).
\[ T = k[(3a - 2b) - x_0] = \frac{1}{a - b}[(3a - 2b) - (3b - 2a)] \] Simplify the numerator:
\[ (3a - 2b - 3b + 2a) = 5a - 5b = 5(a - b) \] \[ T = \frac{1}{a - b} \times 5(a - b) = 5 \, \text{N} \]
Final Answer:
\[ \boxed{5 \, \text{N}} \]