150 Follower Problem!

Simply use virtual work method.

For any close system $\sum { \vec { T } .\vec { V } } =0$

Let the velocity of the block be $v_{0}$ (in horizontal direction).

So $-TV+Tv_{0}cos(90-\theta)=0$

So $v_{0}=300 m/s$

Basic Consideration : Velocity along the string is the same everywhere.

Let the required horizontal velocity be ${V}_{x}$ , then

${ V }_{ x }\cos { { 60 }^{ \circ } } =V$ , or ${ V }_{ x }=300\quad \text{m/s}$ .

Considering H as the height of pulley, R the lenght of the rope and x the distance of the block to the vertical of the pulley, being R=R(t) , x=x(t) and H equal a constant. Now we can write by Phytagoras

$R^{2}=H^{2}+x^{2}$ so derivating by t and after simplification we get

$R\frac{dR}{dt} =x\frac{dx}{dt}$

Since $\frac{R}{x} = \frac{1}{\cos \theta}$

By substitution of given value we get the velocity of the block is twice of the rope

$This\quad figure\quad shows\quad the\quad position\quad of\quad the\quad block\quad after\quad a\\ neglegible\quad time\quad \Delta .\\ \\ Since\quad the\quad velocity\quad of\quad the\quad rope\quad is\quad V,\quad its\quad length\quad\\ on\quad the\quad left\quad side\quad of\quad the\quad pulley\quad changes\quad by\quad \Delta .V\quad in\quad \Delta \quad time.\\ So\quad AB=\Delta .V\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( 1 \right) \\ \\ AC\quad is\quad the\quad displacement\quad covered\quad in\quad \Delta \quad time\quad by\quad the\quad block.\\ \\ So,\quad AC=\Delta .U\quad (Where\quad U\quad is\quad the\quad velocity\quad of\quad the\quad block)\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \left( 2 \right) \\ By\quad simple\quad trigo\quad we\quad can\quad see\quad that\quad AC=AB\csc { \theta } \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left( 3 \right) \\ \\ Using\quad \left( 1 \right) ,\left( 2 \right) \quad and\quad \left( 3 \right) ,\quad we\quad get\\ \Delta .U=\Delta .V\csc { \theta } \\ So\quad U=V\csc { \theta } \\ So\quad U=150\times 2=300$