Question 6: Work, Energy and Power & Circular Motion

Question

Two blocks with masses 100 g and 200 g are attached to the ends of springs $A$ and $B$ as shown in figure. The energy stored in $A$ is $E$. The energy stored in $B$, when spring constants $k_A, k_B$ of $A$ and $B$, respectively satisfy the relation $4 k_A=3 k_B$, is :

Accepted Answer

When a mass $m$ is suspended from a spring of constant $k$, the spring stretches.
If the stretch in spring is $\mathrm{x}, \mathrm{kx}=\mathrm{mg} \Rightarrow \mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}$
The potential energy ( $U$ ) stored in the spring is given by :
$$ \mathrm{U}=\frac{1}{2} \mathrm{kx}^2 $$
Substituting $\mathrm{x}=\frac{\mathrm{mg}}{\mathrm{k}}$ :
$$ \mathrm{U}=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{mg}}{\mathrm{k}}ight)^2=\frac{(\mathrm{mg})^2}{2 \mathrm{k}} $$
Mass of block A is $\mathrm{m}_{\mathrm{A}}=100 \mathrm{~g}$.
The spring constant is $\mathrm{k}_{\mathrm{A}}$.
So, the stored energy in spring A is $\mathrm{E}_{\mathrm{A}}=\mathrm{E}$.
Using the formula :
$$ E=\frac{\left(m_A gight)^2}{2 k_A} $$
Mass of block B is $\mathrm{m}_{\mathrm{B}}=200 \mathrm{~g}=2 	imes 100 \mathrm{~g}=2 \mathrm{~m}_{\mathrm{A}}$
The spring constant of spring B is $\mathrm{k}_{\mathrm{B}}$.
It is given that $4 \mathrm{k}_{\mathrm{A}}=3 \mathrm{k}_{\mathrm{B}} \Rightarrow \mathrm{k}_{\mathrm{B}}=\frac{4}{3} \mathrm{k}_{\mathrm{A}}$.
So, the stored energy in spring B is
$$ \mathrm{E}_{\mathrm{B}}=\frac{\left(\mathrm{m}_{\mathrm{B}} \mathrm{~g}ight)^2}{2 \mathrm{k}_{\mathrm{B}}} $$
Substituting $\mathrm{m}_{\mathrm{B}}=2 \mathrm{~m}_{\mathrm{A}}$ and $\mathrm{k}_{\mathrm{B}}=\frac{4}{3} \mathrm{k}_{\mathrm{A}}$ :
$$ E_B=\frac{\left(2 m_A gight)^2}{2\left(\frac{4}{3} k_Aight)}$$
$\begin{gathered}\Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{4 	imes \frac{3}{8} 	imes\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}ight)^2}{\mathrm{k}_{\mathrm{A}}} \ \Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{12}{8} 	imes \frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}ight)^2}{\mathrm{k}_{\mathrm{A}}} \ \Rightarrow \mathrm{E}_{\mathrm{B}}=\frac{3}{2} 	imes \frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}ight)^2}{\mathrm{k}_{\mathrm{A}}} \ \Rightarrow \mathrm{E}_{\mathrm{B}}=3 	imes\left(\frac{\left(\mathrm{m}_{\mathrm{A}} \mathrm{g}ight)^2}{2 \mathrm{k}_{\mathrm{A}}}ight) \ \Rightarrow \mathrm{E}_{\mathrm{B}}=3 \mathrm{E}\end{gathered}$

Answer

$4 E$

Answer

$2 E$

Answer

$\frac{4}{3} E$

Answer

$3 E$

Work, Energy and Power & Circular Motion

Two blocks with masses 100 g and 200 g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E. The energy stored in B, when spring constants k_A, k_B of A and B, respectively satisfy the relation 4 k_A=3 k_B, is :

Choose the correct answer: