Find Friction Force for 3 kg Block on Inclined Plane

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A 3 kg block is placed on a rough inclined plane at a 30-degree angle, with a coefficient of friction of 0.2 and an upward force of 10 N acting parallel to the plane. The normal reaction force is calculated as 30 cos 30, leading to a friction force of 5.20 N using the formula F = uR. However, the calculation does not account for the upward force of 10 N, which affects the net force acting on the block. The correct friction force is determined to be 5 N. The discussion emphasizes the importance of considering all forces acting on the block for accurate calculations.
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A block of mass 3 kg is on a rough inclined plane at 30 degree to the horizontal. The coefficient of friction is 0.2 and a force of 10 N acts upward parallel to the plane. Find the friction?


The answer to the question is 5N

my work

Normal reaction= 30 cos 30

Using F = u R

(0.2 x 30 cos 30 )= 5.20 N

I am not obtaining the answer, and i don't even know whether the method is correct or not as i have not consider the 10 N. Please need help to attempt the question...!
 
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Take g = 9.8 m/s^2.
 
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