Discussion Forum

1)

A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of a radius of 40 m. 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 the figure, 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 

Answer:

Option A

Explanation:

$N= mg \cos 60^{0}=\frac{mv^{2}}{R}$

  

$\therefore$      $N=\frac{mv^{2}}{R}+\frac{mg}{2}$

  $N=\frac{(1)(10)^{2}}{40}+\frac{(1)(10)}{2}$

  = 7.5 N