A block of mass 10 kg is being pushed along a rough horizontal surface having a coefficient of kinetic friction _k = 0.2 . The applied force is such that the velocity of the block varies with its position x as v = 2x , where v is in m/s and x is in meters. The magnitude of the applied force when the block is at x = 3 m is (Take g = 10 m/s ^2 ) :

Step-by-step solution

The velocity of the block is given by v = 2x . The acceleration a of the block is: a = v dv dx = (2x)(2) = 4x At x = 3 m , the acceleration is: a = 4(3) = 12 m/s ^2 The net force required to produce this acceleration is: F_ net = ma = 10 12 = 120 N The kinetic friction acting on the block is: f_k = _k mg = 0.2 10 10 = 20 N From Newton's Second Law, the net force is the applied force minus the frictional force: F_ net = F_ app - f_k 120 = F_ app - 20 F_ app = 140 N Answer: 140 N