A block of mass 2   kg moving on a horizontal surface with speed of 4   m   s - 1 enters a…

A block of mass 2 kg moving on a horizontal surface with speed of 4 m s-1 enters a rough surface ranging from x=0.5 m to x=1.5 m. The retarding force in this range of rough surface is related to distance by F=-kx where k=12 N m-1. The speed of the block as it just crosses the rough surface will be 
  1. 2 m s-1
  2. 2.5 m s-1
  3. 1.5 m s-1
  4. zero

Solution

Acceleration of the block a=fm=-12x2=-6x (Here the block is retarding)

Now, using, a=vdvdx
⇒v dv = a dx
On integrating, we get

 ∫4vvdv=-6∫0.51.5xdx

⇒v2-422=-61.52-0.522

⇒v2=16-12⇒v=2 m s-1

Asked in: JEE Main 2022 (28 Jun Shift 2)

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