Question

A 1.4 kg block slides across a rough surface such that it slows down with an acceleration of 1.25 m/s(squared). What is the coefficient of kinetic friction between the block and the surface?
Help please. Show steps.

Answer 1

The force of friction is proportional to the normal force of the object. The constant of proportionality is the friction coefficient. In notation
\[ F_{friction} = F_f = \mu N \]
where \( \mu \) = coefficient of friction.

Now, the normal force is the the surface responding to the weight of the object,
\[ Weight = mg = N\]

Using all that, we see that the force of friction is given by
\[ F_f = \mu m g \ \ \ -- (1) \]

On the other hand, friction is the only force acting on the block, hence
\[ F_{net} = F_f = ma \ \ \ -- (2) \]

You know the acceleration rate. Now use equations (1) and (2) to find \( \mu \).