An object slides on a level surface in the +x direction. It slows and comes to a stop with a constant acceleration of -2.45 m/s2. What is the coefficient of kinetic friction between the object and the floor?

uk = 0.25

Explanation:

Given:-

- An object comes to stop with acceleration, a = -2.45 m/s^2

Find:-

What is the coefficient of kinetic friction between the object and the floor?

Solution:-

- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).

- We will apply equilibrium equation on the object in vertical direction.

                               N - m*g = 0

                               N = m*g

Where,  N : Contact force exerted by the surface on the floor

             g : Gravitational acceleration constant = 9.81 m/s^2

- Now apply Newton's second law of motion in the horizontal ( x-direction ):

                             - Ff = m*a

- The frictional force is related to contact force (N) by the following expression:

                              Ff = uk*N

- Substitute the 1st and 3rd expressions in the 2nd equation:

                             uk*m*g = -m*a

                             uk = a / g

- Plug in the values and solve for uk:

                             uk = - (-2.45) / 9.81

                             uk = 0.25        

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