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Sagot :
Sure, let's tackle the problem step by step to find the frictional force.
Step-by-Step Solution:
1. Understanding the physical scenario:
- We know that 40 Joules of energy is expended.
- A classmate is dragged 4 meters across the floor.
- The velocity is constant, implying the applied force and frictional force are equal in magnitude but opposite in direction.
2. Relevant Physics Principle:
- When a constant force moves an object over a distance, the work done (or energy expended) is given by the formula:
[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \][/tex]
- In this case, the work done is due to frictional force since the motion is at a constant velocity.
3. Given Values:
- Work done (energy expended): [tex]\( W = 40 \)[/tex] Joules
- Distance dragged: [tex]\( d = 4 \)[/tex] meters
4. Finding the Frictional Force:
- To find the frictional force ([tex]\( F_{\text{friction}} \)[/tex]), we rearrange the work formula:
[tex]\[ \text{Force} = \frac{\text{Work}}{\text{Distance}} \\ F_{\text{friction}} = \frac{W}{d} \][/tex]
- Substituting the given values:
[tex]\[ F_{\text{friction}} = \frac{40 \, \text{Joules}}{4 \, \text{meters}} = 10 \, \text{Newtons} \][/tex]
5. Conclusion:
- The frictional force acting on the classmate is [tex]\( 10 \)[/tex] Newtons.
Force Diagram:
- To sketch the force diagram:
1. Draw the classmate (represented as a box).
2. Label the forces acting on the classmate:
- A horizontal arrow to the right representing the applied force ([tex]\( F_{\text{applied}} \)[/tex]).
- A horizontal arrow to the left representing the frictional force ([tex]\( F_{\text{friction}} \)[/tex]).
- Vertical arrows representing the gravitational force ([tex]\( F_{\text{gravity}} \)[/tex]) downward and the normal force ([tex]\( F_{\text{normal}} \)[/tex]) upward.
```plaintext
F_normal
↑
│
┌──────┼──────┐
│ │ │
│ │ │
│ (Classmate)
│ │ │
│ │ │
└──────┼──────┘
│
↓
F_gravity
Leftward arrow: F_friction = 10 N
Rightward arrow: F_applied = 10 N
```
Summary:
The frictional force on the classmate is [tex]\( 10 \)[/tex] Newtons, as expending 40 Joules of energy to drag the classmate 4 meters at a constant velocity leads to this result. This frictional force opposes the applied force and balances it out, resulting in constant velocity.
Step-by-Step Solution:
1. Understanding the physical scenario:
- We know that 40 Joules of energy is expended.
- A classmate is dragged 4 meters across the floor.
- The velocity is constant, implying the applied force and frictional force are equal in magnitude but opposite in direction.
2. Relevant Physics Principle:
- When a constant force moves an object over a distance, the work done (or energy expended) is given by the formula:
[tex]\[ \text{Work} = \text{Force} \times \text{Distance} \][/tex]
- In this case, the work done is due to frictional force since the motion is at a constant velocity.
3. Given Values:
- Work done (energy expended): [tex]\( W = 40 \)[/tex] Joules
- Distance dragged: [tex]\( d = 4 \)[/tex] meters
4. Finding the Frictional Force:
- To find the frictional force ([tex]\( F_{\text{friction}} \)[/tex]), we rearrange the work formula:
[tex]\[ \text{Force} = \frac{\text{Work}}{\text{Distance}} \\ F_{\text{friction}} = \frac{W}{d} \][/tex]
- Substituting the given values:
[tex]\[ F_{\text{friction}} = \frac{40 \, \text{Joules}}{4 \, \text{meters}} = 10 \, \text{Newtons} \][/tex]
5. Conclusion:
- The frictional force acting on the classmate is [tex]\( 10 \)[/tex] Newtons.
Force Diagram:
- To sketch the force diagram:
1. Draw the classmate (represented as a box).
2. Label the forces acting on the classmate:
- A horizontal arrow to the right representing the applied force ([tex]\( F_{\text{applied}} \)[/tex]).
- A horizontal arrow to the left representing the frictional force ([tex]\( F_{\text{friction}} \)[/tex]).
- Vertical arrows representing the gravitational force ([tex]\( F_{\text{gravity}} \)[/tex]) downward and the normal force ([tex]\( F_{\text{normal}} \)[/tex]) upward.
```plaintext
F_normal
↑
│
┌──────┼──────┐
│ │ │
│ │ │
│ (Classmate)
│ │ │
│ │ │
└──────┼──────┘
│
↓
F_gravity
Leftward arrow: F_friction = 10 N
Rightward arrow: F_applied = 10 N
```
Summary:
The frictional force on the classmate is [tex]\( 10 \)[/tex] Newtons, as expending 40 Joules of energy to drag the classmate 4 meters at a constant velocity leads to this result. This frictional force opposes the applied force and balances it out, resulting in constant velocity.
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