Question #d4e0c

1 Answer
Apr 1, 2016

Answer:

option (4) i.e#.(muEcostheta)/(sintheta+mucostheta)#

Explanation:

If #h# is the height the object moves up the plane and #l# is the slant distance moved along the plane and #d# is the run.
#h/l=sintheta#

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Force of friction #F_f=muN=mumg cos theta#
Where #mu# is the coefficient of dynamic friction
#m# is the mass
#g# is gravity
#theta# is the angle above horizontal

Work done against friction:

#"work done" = "force due to friction" xx "distance moved "l#
#= (muxxmgcostheta) xx l#

When the object comes to stop, part of its Kinetic energy #E# got converted into its potential energy and rest of it has been used to overcome kinetic friction.

#mgh=E- mumgcos thetaxxl#
#mglsintheta=E- mumgcos thetaxxl#, solving for #l#
#=>l =E/(mgsintheta+mumgcostheta)#
Again
#"work done" = "force due to friction" xx "distance moved "l#
#= (muxxmgcostheta) xx l#
#=(muEmgcostheta)/(mgsintheta+mumgcostheta)= (muEcostheta)/(sintheta+mucostheta)#