Question

Two springs A and B having spring constant K$_A$ and K$_B$ (K$_A$ = 2K$_B$) are stretched by applying force of equal magnitude. If energy stored in spring A is E$_A$ then energy stored in B will be

• A. $2E_A$
• B. $E_A/4$
• C. $E_A/2$
• D. 4$E_A$

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the force applied to a spring, the spring constant, and the energy stored in the spring. The energy (E) stored in a spring when it is compressed or stretched by a displacement (x) is given by the formula:

E = \frac{1}{2} K x^2

Where:

• K is the spring constant.
• x is the displacement from the equilibrium position.

In this question, we have two springs, A and B, with spring constants K_A and K_B respectively, where K_A = 2K_B.

The forces applied to both springs are of equal magnitude. This means:

F = K_A x_A = K_B x_B

Since the springs are stretched by equal forces, we can equate the forces:

K_A x_A = K_B x_B

Given K_A = 2K_B, we can substitute:

2K_B x_A = K_B x_B

Which simplifies to:

2x_A = x_B

Now, we calculate the energy stored in spring B:

The energy stored in spring A is:

E_A = \frac{1}{2} K_A x_A^2

The energy stored in spring B is:

E_B = \frac{1}{2} K_B x_B^2

Substitute x_B = 2x_A:

E_B = \frac{1}{2} K_B (2x_A)^2

Simplify the expression:

E_B = \frac{1}{2} K_B \cdot 4x_A^2

E_B = 2 K_B x_A^2

Substitute K_A = 2K_B into K_A x_A^2 = E_A:

E_A = 2 K_B x_A^2

Thus, E_B = 2E_A.

Therefore, the energy stored in spring B will be twice the energy stored in spring A. Hence, the correct answer is 2E_A.