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Sagot :
To determine the net force on an object that is perpendicular to the surface of an inclined plane, let's break down the steps.
1. Understanding the Forces Acting on the Object:
When an object is on an inclined plane, its weight (gravitational force) acts vertically downward. This gravitational force [tex]\( F_g \)[/tex] can be decomposed into two components:
- A component parallel to the incline [tex]\( F_{g,\parallel} \)[/tex]
- A component perpendicular to the incline [tex]\( F_{g,\perpendicular} \)[/tex]
2. Breaking Down the Gravitational Force:
The gravitational force acting on the object is given by [tex]\( F_g = m \cdot g \)[/tex], where [tex]\( m \)[/tex] is the mass of the object and [tex]\( g \)[/tex] is the acceleration due to gravity.
3. Perpendicular Component of the Gravitational Force:
The component of the gravitational force that is perpendicular to the surface of the incline is responsible for the normal force [tex]\( F_N \)[/tex]. This perpendicular component can be found using the cosine of the angle of inclination [tex]\( \theta \)[/tex]:
[tex]\[ F_{g,\perpendicular} = F_g \cos(\theta) = m \cdot g \cdot \cos(\theta) \][/tex]
Therefore, the net force on the object perpendicular to the surface of the incline is:
[tex]\[ F_{\perpendicular} = m \cdot g \cdot \cos(\theta) \][/tex]
The correct expression for the net force perpendicular to the surface of the incline is:
C. [tex]\( m g \cos (\theta) \)[/tex]
Hence, the answer is option C.
1. Understanding the Forces Acting on the Object:
When an object is on an inclined plane, its weight (gravitational force) acts vertically downward. This gravitational force [tex]\( F_g \)[/tex] can be decomposed into two components:
- A component parallel to the incline [tex]\( F_{g,\parallel} \)[/tex]
- A component perpendicular to the incline [tex]\( F_{g,\perpendicular} \)[/tex]
2. Breaking Down the Gravitational Force:
The gravitational force acting on the object is given by [tex]\( F_g = m \cdot g \)[/tex], where [tex]\( m \)[/tex] is the mass of the object and [tex]\( g \)[/tex] is the acceleration due to gravity.
3. Perpendicular Component of the Gravitational Force:
The component of the gravitational force that is perpendicular to the surface of the incline is responsible for the normal force [tex]\( F_N \)[/tex]. This perpendicular component can be found using the cosine of the angle of inclination [tex]\( \theta \)[/tex]:
[tex]\[ F_{g,\perpendicular} = F_g \cos(\theta) = m \cdot g \cdot \cos(\theta) \][/tex]
Therefore, the net force on the object perpendicular to the surface of the incline is:
[tex]\[ F_{\perpendicular} = m \cdot g \cdot \cos(\theta) \][/tex]
The correct expression for the net force perpendicular to the surface of the incline is:
C. [tex]\( m g \cos (\theta) \)[/tex]
Hence, the answer is option C.
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