Learning Friction #4

Question

A 2 kg box sits on a table where the coefficient of friction is 0.2. A 20 N horizontal force pulls the box to the right. Find the acceleration. 

Answer 1

Hello Syd,

I assume that the coefficient of friction you are given is the coefficient of kinetic friction between the box and the table. Let us take the direction of the applied horizontal force to be the positive x-axis. Once the box is moving, the net force in the x-direction (horizontal direction) will be

F_net = 20N - f_k,

where f_k is the force of kinetic friction on the box. Here,

f_k = μ_kN,

where μ_k is the coefficient of kinetic friction and N is the normal force. The normal force is the force with which the table pushes up on the box, and in this case, the magnitude of the normal force will equal the magnitude of the weight of the box. Thus, f_k = μ_kmg, and so

F_net = 20N - μ_kmg.

But now using Newton's Second Law F_net = ma, we may rewrite the previous equation as

ma = 20N - μ_kmg.

Substituting the given values, we have

(2kg)a = 20N - .2(2kg)(9.8m/s²),

and solving for a, we obtain

a = 8.04m/s²

Hope that helps! Let me know if you need any additional help.

William