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Q. 11 of 75
An object is kept at x = 0, force
on the object is F (constant) from
t = 0, friction coefficient vary as
=Cx
M
x=
=8
u = cx
X
Ratio of maximum value of
friction force and magnitude of
static friction just after the block
stonned​

Answer 1

Explanation:

x=2y

When frictional force equal to F

f1=μMg=cyMg=F

y=cMgF

Then when block stops then displacement is 2y

force of friction is 

fmax=μMg=2y(cMg)

fmax=2F