TS EAMCET 03 May 2019 Shift 2 Solved Paper

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.‌

1

2

mv2‌‌=‌

2mg

9√3

|

3

∫

0

cos‌10‌x‌d‌x|
‌

1

2

mv2‌‌=‌

20

9√3

m×‌

sin‌30∘

10

v2‌‌=‌

2

9√3

‌‌=0.35=0.33m∕s