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If a block moving up an inclined plane at 30a velocity of 5 m/s, stops after 0.5 s,coefficient of friction will be nearly |
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Answer» Initial velocity u=5m/s, final velocity v=0retardation=am/s2,. time=0.5secv=u+at0=5+a×0.5a=-10m/s2mg can be resolved as mgcos30° (acting opposite to normal reaction)and mgsin30°.(acting downwards)(R ) normal reaction=mgcos30°friction f=uR, where u is cofficient of frictionnet force= -(f+ mgsin30°)ma=-(uR+mgsin30)m(-10)=-(u.mgcos30+mgsin30)10m=m(u×√3/2+1/2).g10/g=√3u/2+1/2take g=10, and solving the equationu=1/√3 |
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