Question

Two springs of force constants $300N/m$ (Spring A) and $400N/m$ (Spring B) are joined together in series. The combination is compressed by $8.75cm$. The ratio of energy stored in $A$ and $B$ is $\frac{E_A}{E_B}.$ Then $\frac{E_A}{E_B}$ is equal to :

• A. $\frac{4}{3}$
• B. $\frac{16}{9}$
• C. $\frac{3}{4}$
• D. $\frac{9}{16}$

Answer 1

Answer: (A)

Given: $k_A=300N/m,kB=400N/m$ Let when the combination of springs is compressed by force $F$. Spring $A$ is compressed by $x$. Therefore compression in spring $B$.
$B_x=(8.75-x)cm$
$F=300\times x=400(8.75-x)$
Solving we get, $x=5cm$
$x_B=8.75-5=3.75cm$
$\frac{E_A}{E_B}=\frac{\frac{1}{2}{k}_A{\left(X_A\right)}^2}{\frac{1}{2}{k}_B{\left(X_B\right)}^2}=\frac{300\times {\left(5\right)}^2}{400\times {\left(3.75\right)}^2}=\frac{4}{3}$