The coefficient of friction between the box and the floor is 0.61.
The given parameters;
The normal force on the box is calculated as follows;
Fₙ = 325 + 425sin(35.2)
Fₙ = 569.8 N
The frictional force on the box is calculated as;
[tex]F_k = Fcos(\theta)\\\\F_k = 425 \times cos(35.2)\\\\F_k = 347.225 \ N[/tex]
The coefficient of friction between the box and the floor is calculated as follows;
[tex]F_k = \mu_kF_n[/tex]
[tex]\mu_k = \frac{F_k}{F_n} \\\\\mu_k = \frac{347.225}{569.8} \\\\\mu_k = 0.61[/tex]
Thus, the coefficient of friction between the box and the floor is 0.61.
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The coefficient of kinetic friction between the box and the floor is 0.60.
Given data:
The weight of box is, W = 325 N.
The magnitude of force on the box is, F = 425 N.
The angle with respect to the horizontal is, [tex]\theta = 35.2^{\circ}[/tex].
The frictional force acting on the box provides the linear acceleration to box. Then,
[tex]f=Fcos\theta\\f = 425 \times cos35.2\\f=347.28 \;\rm N[/tex]
Frictional force is also expressed as,
[tex]f = \mu_{k} \times N[/tex]
Here, N is the normal force and its value is, [tex]N = W+Fsin\theta[/tex].
Solving as,
[tex]f = \mu_{k} \times (W+Fsin\theta)\\347.28 = \mu_{k} \times (325+425 \times sin35.2)\\ \mu_{k} = 0.60[/tex]
Thus, the coefficient of kinetic friction between the box and the floor is 0.60.
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