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. Read more

Steps to Solve

1. Identify the forces acting on the box

The forces acting on the box are gravitational force, normal force, and frictional force. The gravitational force can be calculated as $F_g = mg$ where $m$ is the mass of the box and $g$ is the acceleration due to gravity (approximately $9.81 , \text{m/s}^2$).

2. Calculate the gravitational force components

For a box sliding down an incline at an angle $\theta$, we have:

• The force acting parallel to the incline: $$ F_{\text{parallel}} = mg \sin(\theta) $$
• The force acting perpendicular to the incline: $$ F_{\text{normal}} = mg \cos(\theta) $$

3. Determine the frictional force

The frictional force can be calculated using the coefficient of friction $\mu$ and the normal force: $$ F_{\text{friction}} = \mu F_{\text{normal}} $$ Substituting for the normal force: $$ F_{\text{friction}} = \mu mg \cos(\theta) $$

4. Apply Newton's second law

According to Newton's second law, the net force $F_{\text{net}}$ acting on the box is: $$ F_{\text{net}} = F_{\text{parallel}} - F_{\text{friction}} $$ So we have: $$ F_{\text{net}} = mg \sin(\theta) - \mu mg \cos(\theta) $$

5. Calculate the acceleration

Using Newton's second law, $F = ma$, we set the net force equal to the mass times acceleration $a$: $$ ma = mg \sin(\theta) - \mu mg \cos(\theta) $$ Dividing both sides by $m$ gives: $$ a = g \sin(\theta) - \mu g \cos(\theta) $$

6. Final expression for acceleration

Thus, the final expression for the acceleration of the box sliding down the incline is: $$ a = g (\sin(\theta) - \mu \cos(\theta)) $$