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A small mass slides down an inclined plane of inclination θ with the horizontal. The co-efficient of friction is μ= μo x where x is the distance through which the mass slides down and μo is a positive constant.. then the distance covered by the mass before it stops is
a)(2/ μo)tan θ
b)(4/μo)tan θ
c)(1/2μo)tan θ
d)(1/μo)tan θ
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
Thank u very much Mr. Mayank Tanwar.. Your work is greatly appreciated.
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