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Grade 8IIT JEE Entrance Exam

A body of400g slides on a rough horizontal surface. If the frictional force is 3.0, the magnitude of contact force is

Profile image of Vijay
8 Years agoGrade 8
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1 Answer

Profile image of Sarfaraz Ahamed
8 Years ago
The answer is 5 newton
The contact force on the body =(N2 + f2 )1/2
N=Normal Force on the body which is 0.4kg*10m/s2=4Newton
f=frictional force is 3 Newton (given)
therefore Contact Force is (42 + 32)1/2 = 5 Newton

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