Asled is pulled with a horizontal force of 18 n along a level trail, and the acceleration is found to be 0.39 m/s2. an extra mass m = 4.5 kg is placed on the sled. if the same force is just barely able to keep the sled moving, what is the coefficient of kinetic friction between the sled and the trail?

From Newton's second law of motion, it is identified that the net force applied to the object with mass m, will make it move with an acceleration of a. This can be mathematically translated as,
                        F = m x a
To solve for the mass of the sled, we derive the equation above such that,
                        m = F / a
Substituting,
                       m = (18 N) / (0.39 m/s²)
                          m = 46.15 kg

Then, we add to the calculated mass the mass of the extra material.
                      total mass = 46.15kg + 4.5 kg
                       total mass = 50.65 kg
We solve for the normal force of the surface to the object by calculating its weight.
                     F₂ = (50.65 kg)(9.8 m/s²)
                     F₂ = 496.41 N

The force that would allow barely a movement for the object is equal to the product of the normal force and the coefficient of kinetic friction.
                     F = (F₂)(c)
                      c = F/F₂

Substituting,
                      c = 18 N/496.41 N
                       c = 0.0362

ANSWER: c = 0.0362

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