A box weighing 660 N is sliding across a cement floor and the coefficient of sliding friction between the box and the floor is 0.15. If the force pushing the box is 500 N, what is the acceleration of the box?

To find the acceleration of the box, we need to determine the net force acting on it and then apply Newton's second law of motion.

First, let's find the force of friction acting on the box. The force of friction (\(F_f\)) can be calculated using the formula:

\[
F_f = \mu \cdot F_n
\]

where:
- \(\mu\) is the coefficient of friction (0.15),
- \(F_n\) is the normal force.

Since the box is sliding on a level surface, the normal force is equal to the weight of the box, which is 660 N.

Now we can calculate the frictional force:

\[
F_f = 0.15 \cdot 660 \, \text{N} = 99 \, \text{N}
\]

Next, we calculate the net force (\(F_{net}\)) acting on the box. The net force is the difference between the applied force and the frictional force:

\[
F_{net} = F_{applied} - F_f
\]

Substituting in the values we have:

\[
F_{net} = 500 \, \text{N} - 99 \, \text{N} = 401 \, \text{N}
\]

Now, we can apply Newton’s second law (\(F = m \cdot a\)) to find the acceleration (\(a\)). We need to calculate the mass of the box first using the weight:

\[
F = m \cdot g \implies m = \frac{F}{g}
\]

where \(g\) (acceleration due to gravity) is approximately \(9.81 \, \text{m/s}^2\):

\[
m = \frac{660 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 67.14 \, \text{kg}
\]

Now we can find the acceleration:

\[
a = \frac{F_{net}}{m} = \frac{401 \, \text{N}}{67.14 \, \text{kg}} \approx 5.96 \, \text{m/s}^2
\]

Thus, the acceleration of the box is approximately **5.96 m/s²**.