A body of mass 25kg, moving at 3ms-1 on a rough horizontal floor, is at rest after sliding through a distance of 2.50m on the floor. Calculate the coefficient of sliding friction. [g = 10.0ms\(^{-2}\)]
0.09
0.18
0.36
0.54
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Coefficient of sliding friction = F/R
R= 25×10= 250
F= ma= 25×a
From newton third law of motion
V²=U² + 2as
a= V² – U²/2s
a= 3² – 0 /2 × 2.5
a= 9 /5
a=1.8
F=ma = 25 × 1.8
F= 45
Coefficient of sliding friction= F/R
= 45/250
= 0.18
My school. This is how to solve a question, so that when people reads it, they will understand. Take note
My school people don't usually understand what you are solving
Try making your solvings clear
The problem with WAEC is that they set questions without applying wisdom. First of all, the body accelerates. Meaning that there is friction which opposes motion. The friction was not accounted for because they omitted the for that drags the object forward.
Again, coefficient of friction = frictional force/normal reaction.
Here, they only provided an avenue for the coefficient of friction to be calculated with net force. Which is wrong.
Teachers should learn to teach friction by resolving vectors rather than using formulas.
WAEC is messing up big time.
