A small block of mass 100   g is tied to a spring of spring constant 7 . 5   N   m - 1 and…

A small block of mass 100 g is tied to a spring of spring constant 7.5 N m-1 and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad s-1about point A, then tension in the spring is

  1. 0.75 N
  2. 0.25 N
  3. 0.50 N
  4. 1.5 N

Solution

Centripetal force will be provided by the spring force. Let the elongation in the spring be x, then we can write

kx=mω2(r+x)

7.5 x=2.5(0.2+x)

x=0.55=0.1

Therefore, required tension will beT=kx=0.75 N.

Asked in: JEE Main 2023 (06 Apr Shift 1)

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