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