6.62: You need to look up the last table ("energy contents of fuels") of Appendix C of the textbook.
6.65: Review Example 6.9 first.
7.21,7.24: Review Examples 7.4,7.5 first.
Good luck!
Questions can be posted as comments here.
Someone suggested that I give the answers to even numbered questions of the book that I use for homework (with numbers modified by a random number generator). I think it is a fair request, since then you can practice from the book to see if you get the answer right, before working on the actual MP question. So, here are the answers to the selected textbook questions (not the MP questions):
6.16: 1.9E8 N
6.80: 870
6.69: 42 kJ
7.16: 7.0 MJ and 1.0 MJ
7.18: 5.2 J
7.24: 2.3 kN/m
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15 comments:
Thank you Sam ! This helps a lot
I am having trouble with problem number 6.65 Any assistance?
Zak, did you review Example 6.9?
Sam, I swear I can't do anything with problem 6.53. Technically, shouldn't the force be equal to 'a' times 'x' raised to the 3/2? I'm finding it really frustrating, and would appreciate any advise you can give me.
I think that you are missing the coefficient that would be in front of the 'x' when you integrate it. The integral of 'x' is (1/2)*x^2 (one-half x squared), so when you integrate x^(1/2) there must also be a coefficient. Also, since work is the negative integral of the force, don't forget you would need to multiply by '-a', not just 'a.' I hope this helps you!
Thank you so much breanna!
Sam, I am having trouble with homework problem 6.50 because I do not understand how to multiply vectors when they are broken down into three components, or how to find a vector magnitude from its three components (is it the same as if there were only two components? I do not understand).
Hi, Breanna,
in three dimensions,
The dot product of two vectors A=(Ax,Ay,Az) and B=(Bx,By,Bz) is
A.B=Ax*Bx + Ay*By + Az*Bz
Here '.' is used to mean the dot (=inner) product.
The magnitude of A is
sqrt (Ax^2 + Ay^2 + Az^2).
The general definition of the magnitude of a vector (in any dimension) is:
|A| = sqrt (A.A)
for a given vector A.
Hope this helps!
I got the following question from an anonymous student on prob 6.81. That student actually got the correct answer the first time, but maybe the meaning of the question was unclear.
"I don't get the idea behind this problem. What is this value telling me? Can you explain this problem?"
Can any one help out and explain or share her/his opinion?
The value that you get in 6.81 is the work that is done by the floor on the leg, that which stops it. The problem gives two distances: 1) the distance that the leg dropped freely (y1) and 2) the stopping distance (y2). To solve the problem, you can use the fact that delta Energy is equal to the work done on the leg. You know that Energy equals Kinetic Energy plus Potential Energy. At the starting point, there is zero kinetic energy and the potential energy is = mg(y1 + y2). At the end point (when the leg has stopped falling), there is still zero kinetic energy and there is also zero potential energy. Therefore, since delta Energy = Work, Work = -mg(y1+y2). Work also equals the Force applied to the leg by the floor times the stopping distance (and this value is negative since the work is being done in the opposite direction of the movement): -F(y2)= -mg(y1+y2). If you then divide through by the -y2 you get the answer which is F=mg(y1/y2+1). I hope that this makes sense despite the clumsy notation.
Hooray, j. quinn! Those are really high quality comments!
Thank you professor, but I didn't figure this out on my own. My TA explained this problem in class and his logic made sense to me. I actually solved the problem a different way which doesn't as clearly demonstrate the concepts.
That is quite alright with me. If you can summarize it so nicely, then the knowledge is yours already.
I often say to students that no scientists understand any one thing in exactly the same way. They learn some thing and formulate it on their own. Perhaps a similar thing goes on in all things that we do, like a famous saying* by a famous poet: "good poets imitate, and great poets steal."
* I just found out (thanks to google!) that what he actually said was "... Immature poets imitate; mature poets steal; bad poets deface what they take, and good poets make it into something better, ...".
Ya unfortuantly I'm still not understanding this question 6.63 and I'm not a integral or mathmatic type of guy any simpler way of breaking it down?
hi tim, there was no 6.63 in the homework. or are you really doing 6.63? also, at this point, you should look up my solutions at the course web site, and ask questions if any part of them is not clear.
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