Unless you have a diagram to include or any other additional info, I'll assume the block is being pulled by two opposing forces along the horizontal surface.
Horizontally, the block is under the influence of
• one rope pulling in one direction with magnitude 15 N,
• the other rope pulling in the opposite direction with mag. 5 N, and
• friction, opposing the direction of the block's motion, with mag. 3 N.
It stands to reason that the block is accelerating in the direction of the larger pulling force.
(A) By Newton's second law, we have
15 N + (-5 N) + (-3 N) = <em>m</em> (1 m/s²)
where <em>m</em> is the mass of the block. Solve for <em>m</em> :
7 N = <em>m</em> (1 m/s²)
<em>m</em> = (7 N) / (1 m/s²)
<em>m</em> = 7 kg
(B) The friction force is proportional to the normal force, so that if <em>f</em> is the mag. of friction and <em>n</em> is the mag. of the normal force, then <em>f</em> = <em>µ</em> <em>n</em> where <em>µ</em> is the coefficient of friction.
The block does not bounce up and down, so its vertical forces are balanced, which means the normal force and the block's weight (mag. <em>w</em>) cancel out:
<em>n</em> + (-<em>w</em>) = 0
<em>n</em> = <em>w</em>
<em>n</em> = <em>m</em> <em>g</em>
where <em>g</em> = 9.8 m/s² is the mag. of the acceleration due to gravity.
<em>n</em> = (7 kg) (9.8 m/s²)
<em>n</em> = 68.6 N
Then
3 N = <em>µ</em> (68.6 N)
<em>µ</em> = (3 N) / (68.6 N)
<em>µ</em> ≈ 0.044
→ 
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