Question
Question
If W=F×d
, which of the following equations shows work being calculated using the correct units?(1 point)
Responses
113 N=(17.4 J)×(6.51 m)
113 N=(17.4 J)×(6.51 m)
113 J=(17.4 N)×(6.51 m)
113 J=(17.4 N)×(6.51 m)
113 N=(17.4 J)×(6.51 ft)
113 N=(17.4 J)×(6.51 ft)
113 J=(17.4 N)×(6.51 ft)
Answers
GPT-4o mini
To determine which equation shows work calculated using the correct units, we need to recall the formula for work:
\[ W = F \times d \]
Where:
- \( W \) (work) is measured in joules (J)
- \( F \) (force) is measured in newtons (N)
- \( d \) (distance) is measured in meters (m) or other units (like feet) as long as the context is consistent.
Now let's analyze the options:
1. **113 N = (17.4 J) × (6.51 m)**
This equation incorrectly attempts to equate newtons with joules. It does not represent work correctly.
2. **113 J = (17.4 N) × (6.51 m)**
This equation correctly calculates work since it uses newtons for force and meters for distance. This is valid because \( 17.4 N \times 6.51 m = 113 J \).
3. **113 N = (17.4 J) × (6.51 ft)**
Similar to the first option, it incorrectly equates newtons with a product of joules and feet.
4. **113 J = (17.4 N) × (6.51 ft)**
This equation also calculates work, but it uses feet instead of meters. While still valid, it uses an inconsistent unit (feet rather than meters) in the context of standard SI units, but it still gives a correct dimension for work (joules).
So, the two valid options for calculating work correctly are:
- **113 J = (17.4 N) × (6.51 m)** (which is correct, with SI units)
- **113 J = (17.4 N) × (6.51 ft)** (valid, but using non-SI unit for distance)
Given standard practice in physics, the **correct answer would be:**
**113 J = (17.4 N) × (6.51 m)**
\[ W = F \times d \]
Where:
- \( W \) (work) is measured in joules (J)
- \( F \) (force) is measured in newtons (N)
- \( d \) (distance) is measured in meters (m) or other units (like feet) as long as the context is consistent.
Now let's analyze the options:
1. **113 N = (17.4 J) × (6.51 m)**
This equation incorrectly attempts to equate newtons with joules. It does not represent work correctly.
2. **113 J = (17.4 N) × (6.51 m)**
This equation correctly calculates work since it uses newtons for force and meters for distance. This is valid because \( 17.4 N \times 6.51 m = 113 J \).
3. **113 N = (17.4 J) × (6.51 ft)**
Similar to the first option, it incorrectly equates newtons with a product of joules and feet.
4. **113 J = (17.4 N) × (6.51 ft)**
This equation also calculates work, but it uses feet instead of meters. While still valid, it uses an inconsistent unit (feet rather than meters) in the context of standard SI units, but it still gives a correct dimension for work (joules).
So, the two valid options for calculating work correctly are:
- **113 J = (17.4 N) × (6.51 m)** (which is correct, with SI units)
- **113 J = (17.4 N) × (6.51 ft)** (valid, but using non-SI unit for distance)
Given standard practice in physics, the **correct answer would be:**
**113 J = (17.4 N) × (6.51 m)**