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A 55 kg box rest on a flat surface at the store on Earth. On the right side, a rope is attached that is capable of withstanding 425 N of tension.

a. The coefficient of Static Friction for the box on this floor is .85.
b. The coefficient of kinetic friction for the box on this floor is .42

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Final Answer:

a. The coefficient of Static Friction for the box on this floor is 0.85.

b. The coefficient of kinetic friction for the box on this floor is 0.42.

Step-by-step explanation:

The coefficient of static friction (μₛ) can be determined using the formula μₛ = Fₛₜₐₜᵢc/N, where Fₛₜₐₜᵢc is the maximum static frictional force, and N is the normal force. In this scenario, the tension in the rope (425 N) provides the maximum static frictional force when the box is on the verge of moving. The normal force (N) is equal to the weight of the box, which is mg (55 kg × 9.8 m/s²). Plugging in the values, μₛ = 425 / (55 × 9.8) ≈ 0.85.

The coefficient of kinetic friction (μₖ) is calculated using μₖ = Fₖᵢₙₑₜᵢc/N, where Fₖᵢₙₑₜᵢc is the force required to keep the box moving. Given that the tension in the rope is now acting against kinetic friction, Fₖᵢₙₑₜᵢc is equal to the tension. Therefore, μₖ = 425 / (55 × 9.8) ≈ 0.42.

In summary, the coefficients of static and kinetic friction for the box on this floor are 0.85 and 0.42, respectively, indicating the respective frictional properties when the box is stationary and when it's in motion.

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