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Question

A block of mass 4 kg is kept on an inclined plane having an angle of inclination 37 attached with a spring of force constant 20 N/m as shown in the figure. If the coefficient of friction between the block and incline is 0.8, find the range in which the block can be at rest without slipping from its equilibrium position?

A
2.56 m
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B
5 m
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C
10 m
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D
1.25 m
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Solution

The correct option is A 2.56 m
Case 1: Spring is Compressed:
Let the spring be compressed by x1 length.
R is the normal force acting on the block applied by the inclined plane on it.
At equilibrium,
Fy=0
R=mgcos37
Fx=0
mgsin37+fs=kx1
kx1=mgsin37+μsmgcos37
( fs=μsR=μsmgcos37 is limiting value of friction)
20x1=40×35+0.8×40×45
x1=65+810×85
x1=2.48 m

Case 2: Spring under elongation:

Let the spring be elongated by x2 length.
At equilibrium,
Fy=0
R=mgcos37
Fx=0
mgsin37+kx2=fs
kx2=μsmgcos37mgsin37
( fs=μsR=μsmgcos37 is limiting value of friction)
20x2=0.8×40×4540×35
x2=25(3.23)=0.08 m
the range in which the block can be at rest without slipping from its equilibrium position is x1+x2=2.48+0.08=2.56 m


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