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Sagot :
Given data:
* The force applied on the heavy box is,
[tex]F_a=112\text{ N}[/tex]* The weight of the heavy box is,
[tex]w=140\text{ N}[/tex]Solution:
The frictional force acting on the box is,
[tex]\begin{gathered} F_r=\mu\times w \\ \mu\text{ is the coefficient of friction,} \end{gathered}[/tex]The frictional force acting on the heavy box is equal to the applied force as the heavy box is not moving under the action of applied force.
Thus,
[tex]\begin{gathered} F_r=F_a \\ \mu\times w=112 \\ \mu\times140=112 \\ \mu=\frac{112}{140} \end{gathered}[/tex]By simplifying,
[tex]\mu=0.8\text{ }[/tex]Thus, the coefficient of friction between the floor and the box is 0.8.
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