Find answers to your questions and expand your knowledge with IDNLearn.com. Join our community to access reliable and comprehensive responses to your questions from experienced professionals.
Sagot :
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.
### 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.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.