Question

In figure 4.152, lower pulley is free to move in vertical direction only. Block $A$ is given a uniform velocity $u$ as shown, what is velocity of block B as a function of angle $\theta$.

• A. $u\cos \theta$
• B. $\frac{u}{\cos \theta }$
• C. $\frac{u[1+\sin \theta ]}{\cos \theta }$
• D. $\frac{u[1+\cos \theta ]}{\sin \theta }$

Answer 1

The correct option is C $\frac{u[1+\sin \theta ]}{\cos \theta }$

In this type of question, we will apply a concept in which we say that work done by tension is $0$ as force and displacement are in same direction.

$\therefore \sum T_x=0$

Where $x=$ Displacement

$T=$ Tension

Differentiate the above equation we get,

$\sum T.V=0$ where $V=$ Velocity.

As refer to IMAGE 01

Work done $-T\mu$

As refer to IMAGE 02

Work done $=T_1\cos \theta V$

As refer to IMAGE 03

$\therefore T_1+T_1\sin \theta =T$

$T-1=\frac{T}{1+\sin \theta }$

$-T.u+\frac{T}{(1+\sin \theta )}\cos \theta V=0$

$\therefore -T.u+\frac{T}{(1+\sin \theta }\cos \theta V=0$

$V=\frac{u(1+sin\theta )}{\cos \theta }$