A 25 kg box is released on a 27° inclined plane with a coefficient of kinetic friction of 0.3. Calculate the acceleration of the box as it slides down the inclined plane. You must... A 25 kg box is released on a 27° inclined plane with a coefficient of kinetic friction of 0.3. Calculate the acceleration of the box as it slides down the inclined plane. You must draw the FREE-BODY DIAGRAM and follow the procedure discussed in class to receive the answer. Read more

Steps to Solve

1. Identify the forces acting on the box

The box experiences gravitational force ($F_g$), normal force ($N$), and frictional force ($F_f$). The gravitational force can be divided into two components: one parallel to the inclined plane ($F_{g,\parallel}$) and the other perpendicular ($F_{g,\perpendicular}$).

The formulas are:

• Gravitational force: $F_g = m \cdot g$
• $F_{g,\parallel} = F_g \cdot \sin(\theta)$
• $F_{g,\perpendicular} = F_g \cdot \cos(\theta)$

2. Calculate the gravitational force

Given:

• Mass ($m$) = 25 kg
• Acceleration due to gravity ($g$) = 9.81 m/s²

Calculate the gravitational force: $$ F_g = m \cdot g = 25 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 245.25 , \text{N} $$

3. Determine the components of gravitational force

Calculate the parallel and perpendicular components using the angle ($\theta = 27^\circ$):

• Parallel component: $$ F_{g,\parallel} = F_g \cdot \sin(27^\circ) \approx 245.25 , \text{N} \cdot 0.4540 \approx 111.12 , \text{N} $$
• Perpendicular component: $$ F_{g,\perpendicular} = F_g \cdot \cos(27^\circ) \approx 245.25 , \text{N} \cdot 0.8480 \approx 207.82 , \text{N} $$

4. Calculate the frictional force

The frictional force can be calculated using: $$ F_f = \mu_k \cdot N $$ where $\mu_k$ is the coefficient of kinetic friction (0.3), and $N$ is the normal force (equal to $F_{g,\perpendicular}$).

So, $$ F_f = 0.3 \cdot 207.82 , \text{N} \approx 62.35 , \text{N} $$

5. Apply Newton's second law

The net force acting on the box as it slides down the incline is: $$ F_{\text{net}} = F_{g,\parallel} - F_f $$ Now substitute the known values: $$ F_{\text{net}} \approx 111.12 , \text{N} - 62.35 , \text{N} \approx 48.77 , \text{N} $$

Now, use Newton's second law to find the acceleration ($a$): $$ F_{\text{net}} = m \cdot a \implies a = \frac{F_{\text{net}}}{m} = \frac{48.77 , \text{N}}{25 , \text{kg}} \approx 1.95 , \text{m/s}^2 $$