A small block of mass `1` kg is a circular are of ratius `40` m . The block sides along the track without topping and a frictionnal force acts on it in the direction opposite in the instrmens velocity . The work done in evercoming the friction up to the point `Q` as shown is the figure below is `150 J`
(Take the acceleration due to gravity `g = 10 ms^(-2))`

The speed of the block when it reaches the point `Q` is

Options

A. `5 ms^(-1)`
B. `10 ms^(-1)`
C. `10sqrt(3) ms^(-1)`
D. `20 ms^(-1)`

Answer

B

Explanation

Using work energy theorem
`mg R sin 30^@ = W_f = 1/2 mv^2 , " "200 - 150 = (v^2)/2 =< v = 10 ms^(-1)` .