In the figure shown a force F→=(i^+4j^) N acts on a 1 kg block. ,what is the force of friction acting on the block ?

F→=(i^+4j^) N  and\ m=1kgmg=10 N.If N be the normal reaction , then N+4=mgN+4=10N=6 Nfmax=μN=0.3×6 N=1.8NForce in horizontal direction is 1 N which is less than 1.8 N then force of frictions is −i^