A small block of mass `1kg` is released from rest at the top of a rough track. The track is circular arc of radius `40m`. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in figure. is `150J`. (Take the acceleration due to gravity, `g=10ms^-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^@ = (m v^2)/R , N = 5 + 5/2 = 7.5 N`.