Question

A block of mass 20kg is placed on a rough horizontal plane and a horizontal force of 12N is applied.if coefficient of friction is 0.1 the frictional force acting on it is?

Answer 1

Answer:

I got:
μs=0.51
F=78.4N

Explanation:

At the start the 100N force is just enough to overcome static friction so we can write:
Force=Static Friction
F=μsN
where μs is the coefficient of static friction and N= Normal Reaction that in an horizontal case such this will be equal to the weight of the block, so N=mg.

We get:

F=μs⋅mg
in numbers:
100=μs⋅20⋅9.8
μs=10020⋅9.8=0.51

When the movement starts, kinetic friction kicks in and we have that to have uniform motion we need acceleration equal to zero (constant velocity).

We use Newton's Second Law: Σ→F=m→a
or in our case:
Force−Kinetic Friction=mass⋅acceleration

or

F−μkN=0 because acceleration has to be zero.
F−μk⋅mg=0

in numbers:
F−0.4⋅20⋅9.8=0
F=78.4N

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