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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)(1/2 μo)tan θ d)(1/ μo)tan θ

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 

Grade:11

2 Answers

mayank tanwar
15 Points
8 years ago

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

 

Subho Bhattacharya
33 Points
8 years ago

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

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