JEE Advanced2013PhysicsLaws of MotionActual
Paragraph: 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 radius 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 below, is 150 ~J . (Take the acceleration due to gravity, g
Options
- A7 . 5   N
- B8 . 6   N
- C11 . 5     N
- D225   N
Correct answer
A. 7 . 5   N
Step-by-step solution
From FBD N - m g sin 30 ° = m v 2 R ⇒ N = 1 × 10 2 40 + 1 × 10 2 ⇒ N = 7 .5  N