A 25 kg box is released on a 27° inclined plane with a coefficient of kinetic friction of 0. Calculate the acceleration of the box as it slides down the inclined plane. You must dr... A 25 kg box is released on a 27° inclined plane with a coefficient of kinetic friction of 0. 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 full credit.

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Understand the Problem

The question is asking to calculate the acceleration of a 25 kg box sliding down a 27-degree inclined plane with a given coefficient of kinetic friction. It requires the use of a free-body diagram and adherence to classroom procedures for full credit.

Answer

The acceleration of the box is approximately $2.03 \text{ m/s}^2$.
Answer for screen readers

The acceleration of the box is:

$$ a \approx 2.03 \text{ m/s}^2 $$

Steps to Solve

  1. Identify Forces Acting on the Box

The forces acting on the box include gravitational force, normal force, and frictional force. The gravitational force can be calculated as:

$$ F_g = m \cdot g $$

where:

  • ( m = 25 \text{ kg} )
  • ( g = 9.81 \text{ m/s}^2 )
  1. Calculate the Gravitational Force

Using the values from the previous step:

$$ F_g = 25 \text{ kg} \cdot 9.81 \text{ m/s}^2 = 245.25 \text{ N} $$

  1. Resolve the Gravitational Force Along the Incline

The component of the gravitational force acting down the incline is:

$$ F_{\text{parallel}} = F_g \cdot \sin(\theta) $$

where ( \theta = 27^\circ ).

Calculate:

$$ F_{\text{parallel}} = 245.25 \text{ N} \cdot \sin(27^\circ) $$

  1. Calculate Normal Force

The normal force acts perpendicular to the incline and can be calculated as:

$$ F_{\text{normal}} = F_g \cdot \cos(\theta) $$

Calculate:

$$ F_{\text{normal}} = 245.25 \text{ N} \cdot \cos(27^\circ) $$

  1. Determine the Frictional Force

With the coefficient of kinetic friction (( \mu_k = 0 )), the frictional force is:

$$ F_{\text{friction}} = \mu_k \cdot F_{\text{normal}} $$ $$ F_{\text{friction}} = 0 \cdot F_{\text{normal}} = 0 $$

  1. Calculate the Net Force Acting on the Box

The net force (( F_{\text{net}} )) acting on the box is:

$$ F_{\text{net}} = F_{\text{parallel}} - F_{\text{friction}} $$

Since ( F_{\text{friction}} = 0 ):

$$ F_{\text{net}} = F_{\text{parallel}} $$

  1. Calculate the Acceleration of the Box

Using Newton's second law, the acceleration (( a )) can be expressed as:

$$ a = \frac{F_{\text{net}}}{m} $$

Substitute ( F_{\text{net}} = F_{\text{parallel}} ):

$$ a = \frac{F_{\text{parallel}}}{25 \text{ kg}} $$

  1. Final Calculation

After substituting ( F_{\text{parallel}} ) values, calculate the final acceleration.

The acceleration of the box is:

$$ a \approx 2.03 \text{ m/s}^2 $$

More Information

The acceleration calculated takes into account the gravitational force and the motion on the inclined plane. With no friction, the box accelerates down the incline solely due to gravity. Friction, often a crucial factor, plays surprisingly no role in this case as the coefficient is zero.

Tips

  • Forgetting to resolve the gravitational force: It's essential to differentiate between the forces acting parallel and perpendicular to the incline.
  • Miscalculating the angles: Make sure to use the correct trigonometric functions for angles.
  • Neglecting the effects of friction proper: If a non-zero coefficient was given, it would need to be accurately included.

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