Q52 / 60 · JEE Advanced 2013 (Paper 2) · Physics

Paragraph: A small block of mass $1 \mathrm{~kg}$ is released from rest at the top of a rough track. The track is a circular arc of radius $40 \mathrm{~m}$. 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 $150 \mathrm{~J}$. (Take the acceleration due to gravity, $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ ). Figure

Question: The magnitude of the normal reaction that acts on the block at the point $Q$ is

  • Correct answer
    A
    7.5 N
  • B
    8.6 N
  • C
    11.5  N
  • D
    225 N

Solution

From FBD
N-mgsin30°=mv2R⇒N=1×10240+1×102⇒N=7.5 N

Hence, the correct answer is option: 'A'.

Relevant Questions