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:
\(\[
\text{Work} = \text{Force} \times \text{Distance}
\]\)
- 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): \(\( W = 40 \)\) Joules
- Distance dragged: \(\( d = 4 \)\) meters
4. Finding the Frictional Force:
- To find the frictional force (\(\( F_{\text{friction}} \)\)), we rearrange the work formula:
\(\[
\text{Force} = \frac{\text{Work}}{\text{Distance}} \\
F_{\text{friction}} = \frac{W}{d}
\]\)
- Substituting the given values:
\(\[
F_{\text{friction}} = \frac{40 \, \text{Joules}}{4 \, \text{meters}} = 10 \, \text{Newtons}
\]\)
5. Conclusion:
- The frictional force acting on the classmate is \(\( 10 \)\) 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 (\(\( F_{\text{applied}} \)\)).
- A horizontal arrow to the left representing the frictional force (\(\( F_{\text{friction}} \)\)).
- Vertical arrows representing the gravitational force (\(\( F_{\text{gravity}} \)\)) downward and the normal force (\(\( F_{\text{normal}} \)\)) 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 \(\( 10 \)\) 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.
