Question

Two springs a and b (ka=2kb) are stretched by applying forces of equal magnitudes at the ends if the energy stored in a is e then energy stored b is

Answer 1

Answer:

• Energy (U) Stored in Spring B is 2E Joules.

Given:

1. Relation of Spring Constants $\large{\tt K_a = 2K_b}$.
2. Energy stored in the Spring "A" = E.

Explanation:

$\rule{300}{1.5}$

From the Spring Energy Formula,

$\large\bigstar \: {\boxed{\tt U = \dfrac{1}{2}Kx^2}}$

$\bold{Here}\begin{cases}\text{U Denotes Potential Energy} \\ \text{K Denotes Spring Constant} \\ \text{x Denotes Displacement}\end{cases}$

$\large{\boxed{\tt U = \dfrac{1}{2}Kx^2}}$

Now,

$\large{\tt \hookrightarrow U \propto Kx^2}$

I.e. Spring Potential Energy is Directly proportional to Spring constant & Displacement.

Now,

$\large{\tt \hookrightarrow \dfrac{U_a}{U_b} = \dfrac{K_ax^2}{K_bx^2}}$

• $\large{\tt U_a = E}$
• $\large{\tt U_b = U}$

$\large{\tt \hookrightarrow \dfrac{E}{U} = \dfrac{K_ax^2}{K_bx^2}}$

Here the Displacement will be same.

$\large{\tt \hookrightarrow \dfrac{E}{U} = \dfrac{K_a\cancel{x^2}}{K_b\cancel{x^2}}}$

$\large{\tt \hookrightarrow \dfrac{E}{U} = \dfrac{K_a}{K_b}}$

Now, Substituting the values of Spring constants.

$\large{\tt \hookrightarrow \dfrac{E}{U} = \dfrac{K_a}{2K_a}}$

$\large{\tt \hookrightarrow \dfrac{E}{U} = \cancel{\dfrac{K_a}{2K_a}}}$

$\large{\tt \hookrightarrow \dfrac{E}{U} = \dfrac{1}{2}}$

Cross - Multiplying it,

$\large{\tt \hookrightarrow U \times 1 = 2 \times E}$

$\large{\tt \hookrightarrow U = 2 \times E}$

$\huge{\boxed{\boxed{\tt U = 2E \: J}}}$

∴ Energy (U) Stored in Spring B is 2E Joules.

$\rule{300}{1.5}$