Question

Two spring force constant 300 N/m (spring A) and 400 N/m (spring B) are joined together in series. The combination is compressed by 8.75 cm. The ratio of energy stored in A and B is $E_{A/}{E}_B$. is equal to :

• A. 4/3
• B. 16/9
• C. 3/4
• D. 9/10

Answer 1

The correct option is A 4/3

Given that,

$k_A=300N/m$

$k_B=400N/m$

Let the combination of springs is compressed by a force F. Spring A is compressed by x so the compression in spring B is

$x_B=(8.75-x)$

From Newton’s law:

$k_A{x}_A=k_B{x}_B$

$300\times x=400(8.75-x)$

$x=5cm$

$x_B=8.75-5=3.75cm$

So the ratios of energies is:

$\frac{E_A}{E_B}=\frac{1/2k_A{x}_A^2}{1/2k_B{x}_B^2}$

$\frac{E_A}{E_B}=\frac{300\times 5^2}{400\times {3.75}^2}$

$\frac{E_A}{E_B}=\frac{4}{3}$