A small mass slides down an inclined plane of inclinationθ with the horizontal. The co-efficient of friction isμ=μo x where x is the distance through which the mass slides down andμo is a positive constant.. then the distance covered by the mass before it stops isa)(2/μo)tanθb)(4/μo)tanθc)(1/2μo)tanθd)(1/μo)tanθ

THE ACC. OF THE BLOCK IS A=MGSIN@-F/M

A= MGSIN@ - kMGCOS@

VdV/dX=MGSIN@-kmgcos@

integrating with proper limits

vdv/dx=[mgsin@-kmgcos@]

                0=[mgsin@-kmgcos@]dx

putting k=k0x              

mgsin@x=k0mgcos@x*x/2

mgsin@ =k0mgcos@x/2

k0x/2      =tan@

in the solution i have replaced by k0

Approved

Thank u very much Mr. Mayank Tanwar.. Your work is greatly appreciated.