A block of mass 1 kg is initially at rest on a rough horizontal surface having a coefficient of kinetic friction _k = 0.4 . It is pushed along the surface by a variable horizontal force. The applied force decreases linearly with position x from 10 N at x = 0 to 0 N at x = 10 m . The maximum kinetic energy attained by the block during its motion is: (Take g = 10 m/s ^2 )

Step-by-step solution

The kinetic friction acting on the block is f_k = _k mg = 0.4 1 10 = 4 N . The applied force is F(x) = 10 - x . The net force on the block is F_ net = F(x) - f_k = 10 - x - 4 = 6 - x . The kinetic energy of the block will be maximum when its acceleration becomes zero, which occurs when the net force is zero. F_ net = 0 6 - x = 0 x = 6 m . The maximum kinetic energy is equal to the net work done on the block up to x = 6 m . W_ net = ₀⁶ F_ net , dx = ₀⁶ (6 - x) , dx W_ net = [ 6x - x^2 2 ]₀⁶ = ( 36 - 36 2 ) = 18 J .