A block of mass 2 kg moving on a horizontal surface with an initial speed of 6 m s ⁻¹ enters a rough region at x = 0 . The retarding force acting on the block in this region is given by F = - x , where is a positive constant and x is the position in meters. If the block comes to rest exactly at x = 9 m , the value of (in N m ^ -1/2 ) is

Step-by-step solution

According to the work-energy theorem, the net work done on the block equals its change in kinetic energy. W = K W = ₀⁹ F , dx = ₀⁹ (- x ) , dx W = - [ x^ 3/2 3/2 ]₀⁹ = - ( 2 3 ) (9^ 3/2 - 0) W = - ( 2 3 ) (27) = -18 The change in kinetic energy is: K = K_f - K_i = 0 - 1 2 m v₀^2 K = - 1 2 (2) (6)^2 = -36 J Equating the work done to the change in kinetic energy: -18 = -36 = 2 Answer: 2