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 magnitude of the normal reaction that acts on the block at the point `Q` is

Options

A. ` 7.5 N`
B. `8.6 N`
C. 11.5 N`
D. `22.5 N`

Answer

A

Explanation

`N - mg cos 60^(@) = (mv^(2))/(t)`
` :. N = mg cos 60^(@) = (mv^(2))/(t) ` …(i)

Loss in P.E ` = mg + 40 sin 30^(@) = 200 J`
work done in over coming friction ` = 1.50 J`
:. K.E. posessed by the particle ` = 30 J`
`:' (1)/(2) mv^(2) = 50 J `
`:' mv^(2) = 100 J ` ...(2)
from (1) and (2) , `N = 1 xx 10 xx(1)/(2) +(100)/(40) = 5 +2.5 = 7.5 N`
(a) is the correct option