Question

Block $A$ is moving horizontally with velocity $v_A$. Find the velocity of block $B$ at the instant as shown in the figure.

Answer 1

Let $l$ be length of the inclined rope.

$l=\sqrt{h^2+x^2}$.

For every, $t$ amount of rope increase in $l$, B moves $\frac{t}{2}$ distance.

$\Rightarrow V_B=\frac{1}{2}\cdot \frac{dl}{dt}$

Now , $l^2=x^2+h^2$

$\Rightarrow 2l\frac{dl}{dt}=2x\frac{dx}{dx}=x\cdot V_A$

$\Rightarrow \frac{dl}{dt}=\frac{x{V}_A}{\sqrt{x^2+h^2}}$

$\Rightarrow V_B=\frac{x}{2}\frac{V_A}{\sqrt{x^2+h^2}}$