Question

Given that, mass of A is 30kg, mass of B is 5kg, and the system is released from rest, and $\textbf{at a certain instant,}$the velocity attained by A is 2.5m/s, find the downward displacement of A and the upward displacement of B $\textbf{at that instant}$
$\sf{\\}$
Kindly solve using $\textsf{Conservation of Mechanical Energy}$​

Answer 1

Applying constraint equation in the main string passing through the two pulleys,

$\sf{\longrightarrow -a_2+2a_1=0}$

$\sf{\longrightarrow a_2=2a_1}$

Since the acceleration is independent of time (constant acceleration) and system is released from rest,

$\sf{\longrightarrow v_2=2v_1}$

and,

$\sf{\longrightarrow s_2=2s_1}$

Applying energy conservation on the system,

$\sf{\longrightarrow U_{iA}+K_{iA}+U_{iB}+K_{iB}=U_{fA}+K_{fA}+U_{fB}+K_{fB}}$

$\sf{\longrightarrow m_1gs_1+0+0+0=0+\dfrac{1}{2}\,m_1(v_1)^2+m_2gs_2+\dfrac{1}{2}\,m_2(v_2)^2}$

$\sf{\longrightarrow m_1gs_1=\dfrac{1}{2}\,m_1(v_1)^2+2m_2gs_1+2m_2(v_1)^2}$

$\sf{\longrightarrow s_1=\dfrac{(m_1+4m_2)(v_1)^2}{2(m_1-2m_2)g}}$

$\sf{\longrightarrow s_1=\dfrac{(30+4\times5)(2.5)^2}{2(30-2\times5)10}}$

$\sf{\longrightarrow\underline{\underline{s_1=0.78125\ m}}}$

And,

$\sf{\longrightarrow\underline{\underline{s_2=1.56250\ m}}}$