Block A of mass 35 kg is resting on a frictionless flier.Another blockB of mass 7kg is resting on it.the coefficient of friction between the blocks is 0.5 while kinetic friction is 0.4if a force of 100 N is applied to block B,the acceleration of the block A will be
Answers
block A .8 metre per second
block B 10.28 metre per second
Explanation:
first we assume that the 7 kg and 35 kg masses are moving with same acceleration .So the net force acting on them is equal to the hundred Newton and net mass of them is 42 kilogram and by using the formula net force is equal to net mass into acceleration of the system gives the acceleration as 2.3 metre per second and now since there is no relative acceleration for motion between two blocks the static friction must act on block a and it is only the force which act on block a and now the force acting on a is find by multiplying 35 kg with 2.3 metre per second and which gives the answer as 80 Newton but since the value of the static friction from the given question is 35 Newton so so kinetic friction must act and now the question from starting new free body diagrams .A free body diagram of block B in which we consider that there is a relative motion between the blocks so the value of the kinetic friction which is calculated as 28 newton is used and from the free body diagram of block B it is found that the acceleration of block B is 10.2 8 metre per second and now the block B is considered and since we know that the only driving force on the block a is the friction force which is currently kinetic friction and so by using the formula force is equal to mass into acceleration we calculate the value of the acceleration of block A . 8 metre per second
