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# A block of mass 0.1 is held against a wall applying a horizontal force of 5 N on the block . If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is :a) 2.5 Nb) 0.98 Nc) 4.9 Nd) 0.49 N

bharat bajaj IIT Delhi
7 years ago
Normal force = F = 5 N
Coeff. Of friction = 0.5

Frictional force = 0.5 × 5 = 2.5 a)

Thanks
BHARAT BAJAJ
Gopal Dahale
40 Points
5 years ago
the answer is b 0.98N but i dont know how. just to disprove the above answer i am writng this. ithink we should also take the mass of the block(0.1kg) into account in order to solve this problem.Thank you
Gopal Dahale
40 Points
5 years ago
Ff = F*Cf = 5*0.5 = 2.5N The weight of the block is 0.1*9.8 = 0.98N The block isn`t moving which means the forces on it are balanced. The weight = 0.98N is balanced by the friction force so Ff = 0.98N But that block won`t move downard until the weight plus downward force exceeds 2.5N or until the upward force exceeds 2.5 + 0.98 N =3.48N