To find the coefficient of kinetic friction between the box and the floor, we start by using the relationship between the force of friction (\(F_f\)), the normal force (\(F_N\)), and the coefficient of kinetic friction (\(\mu_k\)).
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The equation for the force of friction is: \[ F_f = \mu_k F_N \]
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On a horizontal surface, the normal force \(F_N\) is equal to the weight of the box, which can be calculated using: \[ F_N = mg \] where \(m\) is the mass of the box and \(g\) is the acceleration due to gravity (approximately \(9.81 , \text{m/s}^2\)).
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Calculate the weight of the box: \[ F_N = (12.0 , \text{kg})(9.81 , \text{m/s}^2) = 117.72 , \text{N} \]
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Since the box is being pulled at a constant velocity, the force of friction is equal to the applied force, which is \(38.0 , \text{N}\). So: \[ F_f = 38.0 , \text{N} \]
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Now we can set up the equation: \[ 38.0 , \text{N} = \mu_k (117.72 , \text{N}) \]
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Solve for \(\mu_k\): \[ \mu_k = \frac{38.0 , \text{N}}{117.72 , \text{N}} \approx 0.322 \]
Thus, the coefficient of kinetic friction between the box and the floor is approximately 0.322.