A block of mass m is thrown upwards with some initial velocity as shown in figure. On the block {:("Column I","Column II"),("(A) Net force in horizontal direction","(p) Zero"),("(B) Net force in vertical direction",(q) m(g sin theta+mug cos theta)),("(C) Net force along the plane",(r)m(g sin theta cos theta+mu g cos^(2) theta)),("(D) Net force perpendicular to plane",(s)m(g sin^(2) theta+mug sin theta cos theta)):}

A block of mass m is thrown upwards with some initial velocity as show

1.	A block of mass m is thrown upwards with some initial velocity as shown in figure. On the block {:("Column I","Column II"),("(A) Net force in horizontal direction","(p) Zero"),("(B) Net force in vertical direction",(q) m(g sin theta+mug cos theta)),("(C) Net force along the plane",(r)m(g sin theta cos theta+mu g cos^(2) theta)),("(D) Net force perpendicular to plane",(s)m(g sin^(2) theta+mug sin theta cos theta)):}
Answer» SOLUTION :Force of friction and `mg SIN theta` both are DOWNWARDS. `THEREFORE` Acceleration of the block is `a=(g sin theta+mu g cos theta)` down the plane Now, net force in any direction is equal to, F=m(component of acceleration in that direction)