Question

A massless spring gets elongated by amount x 1 ​ under a tension of 5 N . Its elongation is x 2 ​ under the tension of 7 N . For the elongation of (5x 1 ​ −2x 2 ​ ), the tension in the spring will be,

Answer 1

11 N

Answer 2

The problem involves the application of Hooke's Law, which states that the force applied to a spring is directly proportional to its elongation, provided the elastic limit is not exceeded.

Mathematically, Hooke's Law is expressed as:

$F = kx$

where:

$F$ is the tension (force) applied to the spring $x$ is the elongation of the spring $k$ is the spring constant, a measure of the stiffness of the spring. For a given spring, $k$ is constant.

Given information:

1. When the tension is 5 N, the elongation is $x_1$. From Hooke's Law:

   $5 = kx_1 \quad \text{(Equation 1)}$

2. When the tension is 7 N, the elongation is $x_2$. From Hooke's Law:

   $7 = kx_2 \quad \text{(Equation 2)}$

From Equation 1, we can express $x_1$ in terms of $k$:

$x_1 = \frac{5}{k}$

From Equation 2, we can express $x_2$ in terms of $k$:

$x_2 = \frac{7}{k}$

We need to find the tension ($F_{new}$) for an elongation of $(5x_1 - 2x_2)$. Let the new elongation be $x_{new} = (5x_1 - 2x_2)$.

Substitute the expressions for $x_1$ and $x_2$ into the expression for $x_{new}$:

$x_{new} = 5 \left(\frac{5}{k}\right) - 2 \left(\frac{7}{k}\right)$

$x_{new} = \frac{25}{k} - \frac{14}{k}$

$x_{new} = \frac{25 - 14}{k}$

$x_{new} = \frac{11}{k}$

Now, to find the tension ($F_{new}$) corresponding to this new elongation, we again use Hooke's Law:

$F_{new} = k \cdot x_{new}$

Substitute the expression for $x_{new}$:

$F_{new} = k \left(\frac{11}{k}\right)$

$F_{new} = 11 \, \text{N}$

The tension in the spring for the elongation of $(5x_1 - 2x_2)$ will be 11 N.