Question Number 103792 by aurpeyz last updated on 17/Jul/20

Answered by Dwaipayan Shikari last updated on 17/Jul/20

v_f ^2 =v_0 ^2 +2as  400=−2a.120  a=−((400)/(240)) (m/s^2 )=−(5/3) (m/s^2 )(negative means deaccleration)  Force on puck=−m.(5/3) N  Is equal tofrictional force f=−μmg (negative depends on sign convention  −m(5/3)=−μmg  μ=(5/(30))=(1/6)=0.1666666

Answered by Dwaipayan Shikari last updated on 17/Jul/20

(1/2)mv^2 =f.d     {Change in kinetic energy=(1/2)m(v^2 −0)=(1/2)mv^2   (1/2)mv^2 =μmgd   {Work done by the frictional force=f.d  μ=(v^2 /(2gd))=((400)/(2400))=(1/6)=0.16666666..

Commented byaurpeyz last updated on 17/Jul/20

thank you Sir. is there any method apart from conservstion of energy that can be used?