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Question

In the figure shown, find the acceleration in m/s2 of the body kept on the rough horizontal surface. Take g=10 m/s2.


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Solution

The FBD of the block can be shown as

From the FBD we have
ΣFy=0
N+Fsin30=10g
N=10050sin30=75 N
And we have
fmax=μsN=0.3×75=22.5 N

Now, total applied horizontal force F=Fcos30=5032=43.3 N
Since, F>fmax, the block will move

Let acceleration of the block be a
FμkN=ma
43.3(0.1)×75=10a
a=43.37.510=3.58 m/s2

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