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 that act on the box are:

• Gravitational force ($F_g$) vertically downward, calculated as $F_g = mg$, where $m$ is the mass and $g$ is acceleration due to gravity (approximately $9.81 , \text{m/s}^2$).

  For the box, the weight is: $$ F_g = 25 , \text{kg} \times 9.81 , \text{m/s}^2 = 245.25 , \text{N} $$

• The normal force ($F_n$) acting perpendicular to the inclined plane.

• The frictional force ($F_f$) acting opposite to the direction of motion, calculated as: $$ F_f = \mu F_n $$ where $\mu$ is the coefficient of kinetic friction.

2. Calculate the normal force

The normal force can be calculated using the formula: $$ F_n = F_g \cos(\theta) $$ For a $27^\circ$ angle: $$ F_n = 245.25 \cos(27^\circ) $$

Calculating this gives: $$ F_n \approx 245.25 \times 0.8480 \approx 207.49 , \text{N} $$

3. Calculate the friction force

Now, substituting $F_n$ into the friction force equation: $$ F_f = \mu F_n = 0.3 \times 207.49 \approx 62.25 , \text{N} $$

4. Calculate the component of gravitational force along the incline

This force component is calculated by the formula: $$ F_{g,\text{parallel}} = F_g \sin(\theta) $$ Thus, $$ F_{g,\text{parallel}} = 245.25 \sin(27^\circ) $$

Calculating this gives: $$ F_{g,\text{parallel}} \approx 245.25 \times 0.4540 \approx 111.06 , \text{N} $$

5. Determine the net force acting down the incline

The net force ($F_{net}$) is given by: $$ F_{net} = F_{g,\text{parallel}} - F_f $$ So, $$ F_{net} = 111.06 - 62.25 \approx 48.81 , \text{N} $$

6. Calculate the acceleration of the box

Finally, using Newton's second law ($F = ma$), we can find the acceleration ($a$): $$ a = \frac{F_{net}}{m} = \frac{48.81}{25} \approx 1.95 , \text{m/s}^2 $$