A small block starts sliding down an inclined plane forming an angle 45^ horizontal. The coefficient of friction , varies with distance s as =c s^2 where, c is a constant of appropriate dimensions, then distance covered by the block before it stops is

Step-by-step solution

Given that, =C s^2 s= distance and C= constant Net force on the block is gathered M g -f=M a M g - M g =M a g - g =a g [ -C s^2 ]=a a= d v d t =v d v d s a d s=v d v gathered From Eqs. (i) and (ii), we get g [ -C s^2 ] d s=v d v Integrating both sides, we get (g ) s-g C s^3 3 = v^2 2 +K For =45^ , we have, ( g 2 ) s- ( C s^3 3 ) g 2 = v^2 2 +K Initially t=0, s=0, u=0 , substituting these, we get k=0 So, So, aligned & g 2 (s- C s^3 3 ) & = v^2 2 & s- C s^3 3 & = v^2 2 2 g aligned The body stops, when v=0 s- C s^3 3