Question

A motor is fixed inside a box which is moving upwards with velocity $5\text{m/s}$. String is winding at the rate $3\text{m/s}$ around the motor shaft. Then the velocity of block $\text{A}$ will be

• A. $2.5\text{m/s}$ downwards
• B. $5\text{m/s}$ downwards
• C. $1\text{m/s}$ downwards
• D. $2\text{m/s}$ downwards

Answer 1

The correct option is C $1\text{m/s}$ downwards


Since strings is winding at a rate $3\text{m/s}$, the velocity of a point on string near the motor will be,

$v=5\text{m/s (upward)}+3\text{m/s (downward)}$

$\Rightarrow v=5-3=2\text{m/s}$ (upward)

Since the string remains taut the velocity of block $\text{A}$ will be as per string constraint.

$\Rightarrow l_1+l_2+l_3=\text{constant}$

Differentially both sides,

$\frac{d{l}_1}{dt}+\frac{d{l}_2}{dt}+\frac{d{l}_3}{dt}=0$

$\Rightarrow (-2\text{m/s})+v_A+v_A=0$

($\because$ length of string $l_1$ is decreasing, while length of $l_2$ and $l_3$ is increasing)

$\Rightarrow 2v_A=2\text{m/s}$

$\therefore v_A=1\text{m/s}$

Thus velocity of block $\text{A}$ is $1\text{m/s}$ downwards.

Thus correct choice is option (c),

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: In this problem box is moving upward}\\ \text{due to winding of string by motor,}\\ \text{thus motor must be winding in opposite}\\ \text{direction. The net velocity of string in}\\ \text{contact of motor is due to motion of box}\\ \text{and rotation of motor.}\end{array}$