Question

A box weighing 347 N is sliding across a cement floor and the coefficient of sliding friction between the box and the floor is 0.2. If the force pushing the box is 500 N, what is the acceleration of the box?

Answer 1

To determine the acceleration of the box, we first need to calculate the frictional force acting on it. The frictional force (\(F_f\)) can be calculated using the formula:

\[
F_f = \mu \times F_n
\]

where:
- \( \mu \) is the coefficient of friction (0.2 in this case).
- \( F_n \) is the normal force, which for a horizontal surface is equal to the weight of the box (347 N).

Calculating the frictional force:

\[
F_f = 0.2 \times 347 \, \text{N} = 69.4 \, \text{N}
\]

Next, we need to find the net force (\(F_{\text{net}}\)) acting on the box. The net force is the difference between the pushing force and the frictional force:

\[
F_{\text{net}} = F_{\text{push}} - F_f
\]

Given the pushing force is 500 N:

\[
F_{\text{net}} = 500 \, \text{N} - 69.4 \, \text{N} = 430.6 \, \text{N}
\]

Now, we can use Newton's second law to find the acceleration (\(a\)) of the box:

\[
F_{\text{net}} = m \cdot a
\]

First, we need to calculate the mass (\(m\)) of the box using its weight:

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

Where \(g \approx 9.81 \, \text{m/s}^2\):

\[
m = \frac{347 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 35.34 \, \text{kg}
\]

Now substituting into \(F_{\text{net}} = m \cdot a\):

\[
430.6 \, \text{N} = 35.34 \, \text{kg} \cdot a
\]

Solving for \(a\):

\[
a = \frac{430.6 \, \text{N}}{35.34 \, \text{kg}} \approx 12.18 \, \text{m/s}^2
\]

Thus, the acceleration of the box is approximately

\[
\boxed{12.18 \, \text{m/s}^2}
\]