Question

A block of mass M equals to 2 kg is balanced under the action force of f against the vertical wall having coefficient of static friction equals to 0.4 and coefficient of kinetic friction equal to 0.3 then the minimum value for f for which block does not slip is

Answer 1

given
mass=2kg
coefficient of static friction=0.4
coefficient of kinetic friction =0.3
fmin=mg/coefficient of static friction
f=2×10/0.4=50N