In the figure shown the velocity of lift is 2ms- 1 upwards while the string is winding on the motor shaft with velocity 2ms- 1 and block A is moving downwards with a velocity of 2ms- 1 , then find out the velocity of block B .

Question

In the figure shown the velocity of lift is 2ms- 1 upwards while the string is winding on the motor shaft with velocity 2ms- 1 and block A is moving downwards with a velocity of 2ms- 1 , then find out the velocity of block B . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-z0td2dfqbprrneuf.jpg" /> (A) 2ms- 1 uparrow (B) 2ms- 1 downarrow (C) 4 m s- 1 downarrow (D) 8 m s- 1 uparrow

Tardigrade Admin · Accepted Answer

Let, vB = velocity of block B v(A/L) = velocity of block A with respect to lift vL = velocity of lift vB = - v(A/L) + vL + v textwinding = - (. vA - vL .) + vL + v textwinding = - (. - 2 - 2 .) + 2 + 2 =8ms- 1

Q. In the figure shown the velocity of lift is $2 m s^{- 1}$ upwards while the string is winding on the motor shaft with velocity $2 m s^{- 1}$ and block $A$ is moving downwards with a velocity of $2 m s^{- 1}$ , then find out the velocity of block $B$ .

$2 m s^{- 1} ↑$

$2 m s^{- 1} ↓$

$4 m s^{- 1} ↓$

$8 m s^{- 1} ↑$

Let, $v_{B} =$ velocity of block $B$
$v_{\frac{A}{L}} =$ velocity of block $A$ with respect to lift
$v_{L} =$ velocity of lift
$v_{B} = - v_{\frac{A}{L}} + v_{L} + v_{winding}$
$= - (v_{A} - v_{L}) + v_{L} + v_{winding}$
$= - (- 2 - 2) + 2 + 2$
$= 8 m s^{- 1}$

Q. In the figure shown the velocity of lift is 2ms−1 upwards while the string is winding on the motor shaft with velocity 2ms−1 and block A is moving downwards with a velocity of 2ms−1 , then find out the velocity of block B .

2ms−1↑

2ms−1↓

4ms−1↓

8ms−1↑