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Question

In the figure shown below, a 4 kg block is compressed by 2 m. The coefficient of friction between the block and the plane is μ=0.8. Then the frictional force (in N) acting on the block is given by


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Solution

The FBD of the block is as shown in the below figure.


From the FBD, at limiting condition we have
Fy=0 (perpendicular to the plane)
N=mgcos37=40×45=32 N
Fx=0 (along the plane)
mgsin37+f=kx...........(1)
Limiting friction
(fmax)=μsN=0.8×32=25.6 N

Let the maximum compression in spring for which block will not slide be xmax
From eq (1),
kxmax=mgsin37+fmax=40×35+25.6=49.6
xmax=49.6k=49.620=2.48 m

As per the question, the spring is compressed by a length 2 m, so it will not slide ( 2 m<2.48 m)

from equation (1), the friction force acting on the block is given by
mgsin37+f=kx f=kxmgsin37
f=20×240×35=16 N

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