Question

In the arrangement shown in the fig., the ends $P$ and $Q$ of an un-stretchable string move downwards with uniform speed $U$. Pulleys $A$ and $B$ are fixed. Mass $M$ moves upwards with a speed?

• A. 2U cosθ
• B. U/cosθ
• C. 2U/cosθ
• D. U cosθ

Answer 1

The correct option is B

U/cosθ

The block will obviously go up and that too vertically up as both sides are pulled with same velocity.

Lets assume block is going up by velocity $v$

Now focus on one string.

Now AB is the string between pulley and block $M$

AB is being pulled by velocity $U$ and if $M$ block is fixed then the string will stretch which cannot happen as ideal strings are non elastic.

Now here since the block is going up with velocity v then its component $Vcos\theta$ will be towards the string.

Now for the block to go up such that the string is to neither stretch or break or slacken so $U$ = $vcos\theta$ $\Rightarrow$ $U$ = $\frac{U}{cos\theta }$

Alternate solution:

From the triangle MNO, using Pythagoras theorem

$M{N}^2+N{O}^2$ = $M{O}^2$

$\therefore x^2+z^2=L^2$

Here $x$ is a constant.

Differentiating the above equation by $t$:

$0+2z\frac{dz}{dt}=2l\frac{dl}{dt}$(as MN distance is always constant so $\frac{dx}{dt}=0$)

$\Rightarrow$ $V_M$ = $\frac{I}{z}.U$

$V_M$ = $\frac{l}{z}U=\frac{U}{\frac{z}{l}}=\frac{U}{cos\theta }(\because cos\theta =\frac{z}{l})$