In the figure shown, a person pulls a light string with a constant speed `u=10m//s`. The other end of the string is tied to a very small block which moves on a smooth horizontal surface. Find the angle `'theta'`when the block 3 leaves the surface . Take `g=10m//s^(2).`

Answer

`theta=(pi)/(4)`

Explanation

By constraint velocity component of block along the string should be `u`.
`rArrv cos theta u or v=u sec theta ….(1)`

from `(1)a=(dv)/(dt)=u sec theta tan theta (d theta)/(dt) ….(2)`

Initially when block is at a large distance `theta ` is a small component of `T` in vertical direction is very small. As block comes nearer and nearer. `T sin theta` increases and `N` decreases.
When `T sin theta =mg` then block just loses contact with the ground .
so `T sin theta = mg ...............(3)`
`T cos theta =ma ...................(4)`
`(3) & (4) rArr`
`a tan theta =g ..........(5)`

also, `x=h cot theta`
`(dx)/(dt)=-h c o s ec^(2) theta (d theta)/(dt)["as x is decreasing" (dx)/(dt)=-v]`
or `(use theta)/(h cosec^(2)theta)=( d theta)/(dt)...` using (1) .......(6)
usint `(2),(5)` and `(6)` we get
`u sec theta tan theta ((u sec theta)/(h cosec^(2)theta))tan theta =g`
putting values of `u,h & g` we get.
`tan^(4)tehta=1rArrtheta=(pi)/(4)`
Ans. `theta=(pi)/(4)`