A block of mass 1 kg is projected with an initial velocity of 6 m/s at x = 0 on a horizontal track. The track has a rough patch from x = 0 to x = 2 m , where the coefficient of kinetic friction varies with position as _k = 0.1x . Beyond x = 2 m , the track is frictionless. On this smooth portion, the block collides with an ideal horizontal spring of spring constant k = 200 N/m . Taking g = 10 m/s ^2 , the maximum com

Step-by-step solution

The frictional force acting on the block in the rough region is variable and given by: f_k = _k mg = (0.1x)(1)(10) = x N The work done by friction as the block moves from x = 0 to x = 2 m is: W_f = ₀² -f_k dx = ₀² -x dx = - [ x^2 2 ]₀² = - 4 2 = -2 J By the work-energy theorem, the total work done on the block equals its change in kinetic energy. When the spring reaches maximum compression x_m , the block's final velocity is zero. W_f + W_s = K_f - K_i -2 + ( - 1 2 k x_m^2 ) = 0 - 1 2 m v₀^2 Substitute the given va