In a process, the number of cycles to failure decreases exponentially with an…
2016
In a process, the number of cycles to failure decreases exponentially with an increase in load. At a load of 80 units, it takes 100 cycles for failure. When the load is halved, it takes 10000 cycles for failure. The load for which the failure will happen in 5000 cycles is ________.
- A.
40.00
- B.
46.02
- C.
60.01
- D.
92.02
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Correct answer: B
Key idea: cycles to failure follow an exponential decay with load, so model N = C·e^{-kL}.
Use the two given data points: at L = 80, N = 100, and at L = 40, N = 10000. These give 100 = C·e^{-80k} and 10000 = C·e^{-40k}.
Divide the second equation by the first to eliminate C: 10000/100 = e^{40k} ⇒ 100 = e^{40k}. Therefore k = ln(100)/40 ≈ 0.11513.
Find C using 100 = C·e^{-80k}. Since e^{-80k} = e^{-2·40k} = 1/10000, we get C = 100·10000 = 1,000,000.
For N = 5000, solve 5000 = 1,000,000·e^{-kL} ⇒ e^{-kL} = 0.005 ⇒ -kL = ln(0.005) ⇒ L = -ln(0.005)/k.
Compute the value: -ln(0.005) ≈ 5.298317, and dividing by k ≈ 0.11513 gives L ≈ 46.02 units.
Answer: Approximately 46.02 units.
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