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. (a) `5ms^-1`
B. (b) `10ms^-1`
C. (c) `10sqrt3ms^-1`
D. (d) `20ms^-1`

Answer

B

Explanation

Using work energy theorem
`mg R sin 30^@+W_f=1/2mv^2`
`200-150=v^2/2v=10ms^-1`