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 = - k x where k = 12 N m - 1 . The speed of the block as it just crosses the rough surface will be

Step-by-step solution

Acceleration of the block a = f m = - 12 x 2 = - 6 x (Here the block is retarding) Now, using, a = v d v d x ⇒ v   d v   =   a   d x On integrating, we get ∫ 4 v v d v = - 6 ∫ 0 . 5 1 . 5 x d x ⇒ v 2 - 4 2 2 = - 6 1 . 5 2 - 0 . 5 2 2 ⇒ v 2 = 16 - 12 ⇒ v = 2   m   s - 1