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Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Calculating Joules
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********************************************* Question: `q014. An athlete's rate of doing work increases more or less steadily from 340 Joules / second to 420 Joules / second during a 6-minute event. How many Joules of work did she do during this time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 420-340=80 joules difference During 6 minutes 80/6 = 13.33 joules/minute
@& The rate is never less than 340 J / s.
That means every second at least 340 J of work are being done.
There are many more than 13.33 Joules of work done in a minute.
Furthermore the question asked for the work of Joules in 6 minutes.
If it isn't completely clear to you how to reason this out, submit a copy of the question using the question form, along with your reasoning, and I'll be glad to take a look.
It's not necessary for your course, but since you know calculus it's also helpful and instructive to think about how this question is related to an integral of work rate with respect to clock time. *@
####I made a careless error in the work above. I think my thinking below is correct, but I wanted you to see my work and let me know if it is correct.
First we need to convert minutes into seconds so 6 minutes * 60 seconds per minute = 360 seconds.
The average rate is (340+420)/2 which gives an average of 380 joules/second. 380 joules/second *360 seconds = 136,800 joules.
I do also see that you could think of this problem as an integral from 0 seconds to 360 seconds of F*d????
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Very good.
However you aren't given information about force or position, so that integral wouldn't work. You have the right idea, though.
If you assume the rate increases linearly during the 360 seconds, the rate would be the linear function
rate(t) = 340 J / s + 2/9 J/s^2 * t.
You would then integral this function from t = 0 to t = 360 seconds. The integral does come out 136 800 Joules.
The key to the integral is that a thin trapezoid of the graph of rate vs. t would have an average altitude that represents the approximate average rate (wiht respect to time) at which work is being done, and a width that represents a time interval, so that the area would represent average rate of change of work with respect to time * change in time = (change in work) / (change in time) * (change in time) = change in work. Adding all these contributions gives the total amount of work done.
Again you aren't required to understand this. I mention it because of your excellent performance in calculus, and the fact that you might find the deeper insight useful.
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