Question

In the arrangement shown in figure, the string is light and inextensible and friction is absent everywhere. Find the speed of both the blocks after the block $\text{A}$ has ascended a height of $1\text{m}$. Given that $m_A=1\text{kg}$ and $m_B=2\text{kg}$. (Take $g=10{\text{m/s}}^2$)

• A. $1.5\text{m/s}$
• B. $1.55\text{m/s}$
• C. $2.58\text{m/s}$
• D. $2.19\text{m/s}$

Answer 1

The correct option is C $2.58\text{m/s}$

Friction is absent. Therefore, mechanical energy of the system will remain conserved.
From constraint relations, speed of both the blocks will be same. Suppose it is $v$. When block A ascends by $1\text{m}$, block B will descend by the same.

Here, gravitational potential energy of $m_B\left(2\text{kg}\right)$ block is decreasing while gravitational potential energy of $m_A\left(1\text{kg}\right)$ block is increasing and kinetic energy of both the blocks is increasing.

So, we can write using energy conservation :
(Loss in potential energy of $2\text{kg}$ block)$=$(Gain in potential energy of $1\text{kg}$ block) $+$(Gain in kinetic energy of both the blocks)
$\therefore m_{B}gh=m_{A}gh+\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{v}^2$
$⟹\left(2\right)\times \left(10\right)\times \left(1\right)=\left(1\right)\times \left(10\right)\times \left(1\right)+\frac{1}{2}\left(1\right)v^2+\frac{1}{2}\left(2\right)v^2$
$⟹1.5v^2=10$
$⟹v=2.58\text{m/s}$
Hence, the calculated speed is $2.58\text{m/s}$