Question

A box weighing 18 N requires a force of 6.0 N to drag it at a constant rate. What is the coefficient of sliding friction?

Answer 1

0.33 . Equation is Force of friction equals normal force times coefficient of friction, so 6=18u. Divide 6 by 18

Answer 2

Answer:

Coefficient of sliding friction is 0.33.

Explanation:

The force of sliding friction is proportional to the normal force. It is given by:

$F_s=\mu _sF_N$

The normal force of the box is equal to its weight as the force in the vertical direction is balanced.

$F_N= 18N$

$F_s= 6.0N$ (given)

Thus, the coefficient of sliding friction is:

$\mu _s=\frac{F_s}{F_N}=\frac{6.0}{18} = 0.33$

Thus, the coefficient of sliding friction is 0.33.