Ideal machine check — work and efficiency A lifting machine raises a 1000 N load through 0.2 m using an effort of 200 N through 1.0 m. Is this machine ideal (100% efficient)?
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ACorrect
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BIncorrect
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CCorrect only if friction is ignored
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DCorrect only if velocity ratio is 10
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EIncorrect because M.A. ≠ V.R. here
Answer
Correct Answer: Correct
Explanation
Introduction / Context:An ideal machine has no losses, so input work equals output work and efficiency equals 100%. This question tests your ability to verify ideality by comparing work in and work out.
Given Data / Assumptions:
- Load = 1000 N; load distance = 0.2 m.
- Effort = 200 N; effort distance = 1.0 m.
- No additional losses specified (we compute purely from given values).
Concept / Approach:Efficiency η = output work / input work. If η = 1 (or 100%), the machine is ideal. Also, for an ideal machine, Mechanical Advantage (M.A.) = Velocity Ratio (V.R.).
Step-by-Step Solution:
Compute output work: W_out = Load * load distance = 1000 * 0.2 = 200 J.Compute input work: W_in = Effort * effort distance = 200 * 1.0 = 200 J.Efficiency: η = W_out / W_in = 200 / 200 = 1 (100%).Check M.A. and V.R.: M.A. = Load / Effort = 1000 / 200 = 5; V.R. = effort distance / load distance = 1.0 / 0.2 = 5; hence M.A. = V.R., consistent with an ideal machine.Verification / Alternative check:Because work in equals work out exactly, there are no implied energy losses. This confirms the ideal assumption from the given numbers.
Why Other Options Are Wrong:
- Incorrect: Conflicts with equal input and output work.
- Correct only if friction is ignored: Already implied by equal works; the data shows no loss.
- Correct only if V.R. = 10: Irrelevant; V.R. here is 5.
- Incorrect because M.A. ≠ V.R.: They are equal in this case.
Common Pitfalls:Mixing up force ratios with work equality; forgetting that efficiency compares energies, not just forces or distances independently.
Final Answer:Correct