Question

At a given instant, block A is moving upwards with velocity of $5m/s$. What will be velocity of block B at that instant ?
(Take upward velocity to be positive.)

• A. $-15m/s$
• B. $15m/s$
• C. $-5m/s$
• D. $-10m/s$

Answer 1

The correct option is A $-15m/s$


Since $l_1,l_2,l_3\&l_4$ are sections of the same string.
$l_1+l_2+l_3+l_4=\text{constant}$
$\Rightarrow \frac{d{l}_1}{dt}+\frac{d{l}_2}{dt}+\frac{d{l}_3}{dt}+\frac{d{l}_4}{dt}=0$
$\Rightarrow V_A+\frac{d{l}_2}{dt}+\frac{d{l}_3}{dt}-V_B=0....\left(1\right)$
$\because \frac{d{l}_2}{dt}$ is the velocity of pulley and it is attached to the same string with block A, velocity of pulley is same as velocity of block A.
i.e $\frac{d{l}_2}{dt}=\frac{d{l}_3}{dt}=5m/s$
$V_A=5m/s$
Substituting in the constraint equation $\left(1\right)$, we get
$5+5+5-V_B=0$
$V_B=15m/s$ in downward direction
Since velocity in upward direction is positive $\therefore V_B=-15m/s$