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An object of mass [tex]$m$[/tex] slides down an incline with angle [tex]$\theta$[/tex]. Which expression shows the net force on the object perpendicular to the surface of the incline?

A. [tex]F_{N^{-}} m g \cos (\theta)[/tex]
B. [tex]F_N - m g \sin (\theta)[/tex]
C. [tex]m g \sin (\theta)[/tex]
D. [tex]m g \cos (\dot{\theta})[/tex]

Sagot :

GinnyAnsweridn
Let's analyze the forces involved when an object of mass [tex]\( m \)[/tex] slides down an inclined plane with an angle [tex]\(\theta\)[/tex].

### Step-by-Step Solution:

1. Identify the Forces Acting on the Object:
- Gravitational Force [tex]\( F_g \)[/tex]: This force acts vertically downward and has a magnitude of [tex]\( mg \)[/tex].
- Normal Force [tex]\( F_N \)[/tex]: This force acts perpendicular to the surface of the incline and is provided by the surface to support the object.
- Component of Gravitational Force Parallel to the Incline: This component is [tex]\( mg \sin(\theta) \)[/tex].
- Component of Gravitational Force Perpendicular to the Incline: This component is [tex]\( mg \cos(\theta) \)[/tex].

2. Focus on the Perpendicular Components:
- The normal force [tex]\( F_N \)[/tex] must balance out the component of the gravitational force that is perpendicular to the incline. Therefore, the normal force is equal in magnitude but opposite in direction to [tex]\( mg \cos(\theta) \)[/tex].

3. Expression for the Net Force Perpendicular to the Surface:
- Since the normal force [tex]\( F_N \)[/tex] balances the component [tex]\( mg \cos(\theta) \)[/tex] exactly, there is no net force perpendicular to the incline if we are only considering the equilibrium position.
- However, when we look for the net force expression perpendicular to the incline, it simplifies to understanding that the normal force [tex]\( F_N \)[/tex] itself is the reaction force to this component, so we equate [tex]\( F_N \)[/tex] to [tex]\( mg \cos(\theta) \)[/tex].

Thus, the correct expression that describes the net force on the object perpendicular to the surface of the incline is:

[tex]\[ F_N = mg \cos(\theta) \][/tex]

### Selecting the Correct Option:
Given the options we have:
A. [tex]\( F_N - mg \cos(\theta) \)[/tex]
B. [tex]\( F_N - mg \sin(\theta) \)[/tex]
C. [tex]\( mg \sin(\theta) \)[/tex]
D. [tex]\( mg \cos(\theta) \)[/tex]

From the explanation above, the correct answer corresponds to:

[tex]\[ \boxed{\text{D}. \; mg \cos(\theta)} \][/tex]

This completes our step-by-step solution for the problem.
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