# A block of mass 20kg is pushed with a horizntal force of 90N. if the coefficient of static and kinetic friction are 0.4 and 0.3 the frictional force acting on the block if g=10 is

2
by aimanakhtar799
why did you report my answer please clarify.

2014-10-02T22:19:09+05:30

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Mass of block = 20kg
weight = 20X10 = 200N (vertically downward)
Normal reaction = 200N (verticaly upward)

static friction = 0.4X200 = 80N
So the block will move and kinetic friction would act.

Kinetic friction = 0.3X200 = 60N
Net force in horizontal direction = 90-60 = 30N
Net force acting in verical direction = 200-200 = 0

OPTIONS;90N,80N,60N,30N
30N
WHERE did the 90 come from?
anyways the answer is 60N
The 90 is the horizontal force acting on it. it came from the information that you had provided. And yes the kinetic friction is 60N What i have solved is the net force acting on the block. And I have clearly written in the 6th line that kinetic friction is 60N.
2014-10-04T02:16:52+05:30

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The question is about frictional force.   The answer is correct.
Maximum Static Frictional force = coefficient of static friction * weight
= 0.4 * 20 kg * 10 m/sec² = 80 Newtons  (static)
Direction of frictional force is horizontal and is opposite to the direction of 90 Newtons
This is the frictional force until the block starts moving.

When the block starts and is moving:
dynamic or kinetic frictional force = 0.3 * weight = 0.3 * 20 kg * 10 m/s²
= 60 Newtons (in motion)
Direction of frictional force is horizontal and is opposite to the direction of 90 Newtons

The solution is perfectly fine!!