Question
The mechanical advantage of an inclined plane is the ratio of the inclined surface to the vertical height. Simple machines with greater mechanical advantage reduce the force required to lift a load. In which case is the effort required the LEAST?
Responses
A raising a 100 kg box to a height of 1 m using an inclined plane of length 5 mraising a 100 kg box to a height of 1 m using an inclined plane of length 5 m
B raising a 100 kg box to a height of 2 m using an inclined plane of length 6 mraising a 100 kg box to a height of 2 m using an inclined plane of length 6 m
C raising a 200 kg box to a height of 1 m using an inclined plane of length 4 mraising a 200 kg box to a height of 1 m using an inclined plane of length 4 m
D raising a 200 kg box to a height of 2 m using an inclined plane of length 8 mraising a 200 kg box to a height of 2 m using an inclined plane of length 8 m
Responses
A raising a 100 kg box to a height of 1 m using an inclined plane of length 5 mraising a 100 kg box to a height of 1 m using an inclined plane of length 5 m
B raising a 100 kg box to a height of 2 m using an inclined plane of length 6 mraising a 100 kg box to a height of 2 m using an inclined plane of length 6 m
C raising a 200 kg box to a height of 1 m using an inclined plane of length 4 mraising a 200 kg box to a height of 1 m using an inclined plane of length 4 m
D raising a 200 kg box to a height of 2 m using an inclined plane of length 8 mraising a 200 kg box to a height of 2 m using an inclined plane of length 8 m
Answers
GPT-4o mini
To determine which case requires the least effort when using an inclined plane, we need to calculate the mechanical advantage (MA) for each scenario and then find the effort required to lift the load.
Mechanical advantage (MA) is defined as:
\[ \text{MA} = \frac{\text{Length of incline}}{\text{Height lifted}} \]
The effort needed can be calculated using:
\[ \text{Effort} = \frac{\text{Weight of load}}{\text{MA}} \]
Let's compute the MA and the effort for each option.
### Option A
- **Weight of box**: 100 kg = \( 100 \times 9.81 = 981 \, \text{N} \)
- **Height lifted**: 1 m
- **Length of incline**: 5 m
- **MA**:
\[ \text{MA} = \frac{5\, \text{m}}{1\, \text{m}} = 5 \]
- **Effort**:
\[ \text{Effort} = \frac{981 \, \text{N}}{5} = 196.2 \, \text{N} \]
### Option B
- **Weight of box**: 100 kg = \( 981 \, \text{N} \)
- **Height lifted**: 2 m
- **Length of incline**: 6 m
- **MA**:
\[ \text{MA} = \frac{6\, \text{m}}{2\, \text{m}} = 3 \]
- **Effort**:
\[ \text{Effort} = \frac{981 \, \text{N}}{3} = 327 \, \text{N} \]
### Option C
- **Weight of box**: 200 kg = \( 200 \times 9.81 = 1962 \, \text{N} \)
- **Height lifted**: 1 m
- **Length of incline**: 4 m
- **MA**:
\[ \text{MA} = \frac{4\, \text{m}}{1\, \text{m}} = 4 \]
- **Effort**:
\[ \text{Effort} = \frac{1962 \, \text{N}}{4} = 490.5 \, \text{N} \]
### Option D
- **Weight of box**: 200 kg = \( 1962 \, \text{N} \)
- **Height lifted**: 2 m
- **Length of incline**: 8 m
- **MA**:
\[ \text{MA} = \frac{8\, \text{m}}{2\, \text{m}} = 4 \]
- **Effort**:
\[ \text{Effort} = \frac{1962 \, \text{N}}{4} = 490.5 \, \text{N} \]
### Summary of Efforts
- **Option A**: 196.2 N
- **Option B**: 327 N
- **Option C**: 490.5 N
- **Option D**: 490.5 N
### Conclusion
The least effort required is in **Option A** (raising a 100 kg box to a height of 1 m using an inclined plane of length 5 m), with an effort of **196.2 N**.
Mechanical advantage (MA) is defined as:
\[ \text{MA} = \frac{\text{Length of incline}}{\text{Height lifted}} \]
The effort needed can be calculated using:
\[ \text{Effort} = \frac{\text{Weight of load}}{\text{MA}} \]
Let's compute the MA and the effort for each option.
### Option A
- **Weight of box**: 100 kg = \( 100 \times 9.81 = 981 \, \text{N} \)
- **Height lifted**: 1 m
- **Length of incline**: 5 m
- **MA**:
\[ \text{MA} = \frac{5\, \text{m}}{1\, \text{m}} = 5 \]
- **Effort**:
\[ \text{Effort} = \frac{981 \, \text{N}}{5} = 196.2 \, \text{N} \]
### Option B
- **Weight of box**: 100 kg = \( 981 \, \text{N} \)
- **Height lifted**: 2 m
- **Length of incline**: 6 m
- **MA**:
\[ \text{MA} = \frac{6\, \text{m}}{2\, \text{m}} = 3 \]
- **Effort**:
\[ \text{Effort} = \frac{981 \, \text{N}}{3} = 327 \, \text{N} \]
### Option C
- **Weight of box**: 200 kg = \( 200 \times 9.81 = 1962 \, \text{N} \)
- **Height lifted**: 1 m
- **Length of incline**: 4 m
- **MA**:
\[ \text{MA} = \frac{4\, \text{m}}{1\, \text{m}} = 4 \]
- **Effort**:
\[ \text{Effort} = \frac{1962 \, \text{N}}{4} = 490.5 \, \text{N} \]
### Option D
- **Weight of box**: 200 kg = \( 1962 \, \text{N} \)
- **Height lifted**: 2 m
- **Length of incline**: 8 m
- **MA**:
\[ \text{MA} = \frac{8\, \text{m}}{2\, \text{m}} = 4 \]
- **Effort**:
\[ \text{Effort} = \frac{1962 \, \text{N}}{4} = 490.5 \, \text{N} \]
### Summary of Efforts
- **Option A**: 196.2 N
- **Option B**: 327 N
- **Option C**: 490.5 N
- **Option D**: 490.5 N
### Conclusion
The least effort required is in **Option A** (raising a 100 kg box to a height of 1 m using an inclined plane of length 5 m), with an effort of **196.2 N**.