The following figure shows the force acting on the bar.
The minimum force to move the block is given by Fmin=‌
mg
9
‌quad......(1) The force acting on the bar is given by, FH‌‌=F‌cos‌θ ‌‌=F‌cos‌10‌x Here, θ=cx and c=10∘m−1,θ=30∘ Hence, at an angle 30∘ θ=10x⇒30∘=10x x=3m........(2) At 30∘, block start moving. From equation (1) F‌cos‌30∘‌‌=‌
mg
9
F‌‌=‌
2mg
9√3
............(3) From the law of conservation of energy, ‌
1
2
mv2=|∫F⋅cos‌θ‌d‌x| Substitute the values in the above equation.‌