Question

A block of 200N resting on rough horizontal surface is pulled by force 100N 30^(@) to the horizontal.If mu=0.175 frictional force is​

Answer 1

Answer:

According to question,

As shown in below figure

So,The force on the block long the horizontal is

F

x

=100cos30−f⋯(i)

f=μ×N⋯(ii)

and also,

200=N+100sin30

N=150N

Now,

substituting the above value in (ii)

f=150×μ

substituting the value of

′

f

′

in (i)

F

x

=100cos30−150×μ

Hence, the block moves with the constant velocity it means that there is no net force acting on it is horizontal direction,

F

x

=0

100cos30−150×μ=0

86.6025=150×μ

μ=0.57735

Explanation:

here your answer hope it will help you

Answer 2

Answer:

494.87

FS=uN

u=0.175

N=200-100sin30