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A box of mass 20 kg is at rest on a rough horizontal floor. The coefficient of friction...
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Refer to the attached image for the given situation.
`rArr F_(norm)=F_(grav)=m*g=20*9.81=196.2 N`
Again, along the x-direction,
F_(fric)=coefficient of friction*F_(norm)
Acceleration, a = F_(net)/m
Therefore, the frictional force on the surface is 58.86 N and magnitude of the acceleration of the box is 0.307 m/s^2.
Posted by llltkl on August 3, 2013 at 3:19 PM (Answer #1)
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