Final answer:
The magnitude of the force of friction when an 89 kg refrigerator is sliding with a coefficient of kinetic friction of 0.446 is approximately 389.04 N. The answer options provided do not match this calculation, suggesting there may be an error in the choices given.
Step-by-step explanation:
The question asks for the magnitude of the force of friction when an 89 kg refrigerator is sliding on the ground, given that the coefficient of kinetic friction (μk) is 0.446. To calculate the force of friction, we can use the formula:
f_k = μk * N
where N is the normal force, which equals the weight of the refrigerator (mass times gravity, m*g) in this case, since there are no other vertical forces acting on it. With Earth's gravity at approximately 9.8 m/s², the calculation is as follows:
f_k = 0.446 * (89 kg * 9.8 m/s²)
f_k = 0.446 * 872.2 N
f_k = 389.04 N
After performing the calculation, we find that the magnitude of the force of friction is approximately 389.04 N. This is not one of the answer choices, so we should double-check the calculation. Since there is no choice that matches our answer and we trust our calculations, the question might contain incorrect answer choices, or the provided coefficient of kinetic friction may not be accurate for this scenario.