Inclined Planes, finding final velocity

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Homework Help Overview

The discussion revolves around a problem involving inclined planes and the calculation of final velocity using energy conservation principles. The original poster attempts to find the final velocity of an object sliding down an incline, using the height derived from the incline's hypotenuse and applying the energy conservation equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the energy conservation equation, questioning the correctness of the textbook's answer compared to the calculated result. Some participants express skepticism about the accuracy of the textbook, suggesting it may contain errors.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's calculations. There is a recognition of potential discrepancies in the textbook's answer, and some participants are gathering information to identify patterns of errors in the textbook.

Contextual Notes

Participants note the importance of the textbook's title and publisher for tracking potential errors, as well as the original poster's previous inquiries regarding similar problems. There is an acknowledgment of the original poster's efforts to verify their understanding against the textbook's claims.

ericcy
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Homework Statement
A 1975kg car is parked at the top of a steep 42m long hill inclined at an angle of 15 degrees. If the car starts rolling down the hill, how fast will it be going when it reaches the bottom of the hill (neglect friction).
Relevant Equations
Ei=Ef, Ek=1/2mv^2, Ep=mgh
I determined 42m to be the hypotenuse so I used sine law to find the height of the incline, 10.87m. I used this height in the equation Ei=Ef, since they should be equal.

Ei=Ef
mgh=1/2mv^2 (at the start there is no kinetic energy, at rest. at the end there is only kinetic, no potential)
1975(9.81)(10.87)=1/2(1975)v^2
210603.532/987.5=987.5v^2/987.5
(square root)213.269=(square root)v^2
V=14.603

Answer in the back of the book is 0.4m/s, for whatever reason. All responses are appreciated, thanks.
 
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Your result looks good to me. Assuming that the statement of the problem is a full and correct representation of the textbook's version, then I suspect that the book's answer key may be mistaken for this problem.
 
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@ericcy, can you let us know the title and publisher of your textbook so that we Mentors and Homework Helpers can keep an eye out for similar issues?

If we see a pattern of such typos or content errors related to a given text we can often shorten the time-to-resolution of questions where the calculated results don't match the text's answers. Thaks for your help!
 
gneill said:
@ericcy, can you let us know the title and publisher of your textbook so that we Mentors and Homework Helpers can keep an eye out for similar issues?

If we see a pattern of such typos or content errors related to a given text we can often shorten the time-to-resolution of questions where the calculated results don't match the text's answers. Thaks for your help!
Turns out for this one I was just looking in the wrong section, lol. But there are quite a few mistakes in the back of the book that I've noticed. That's kind of why I've been asking a lot of questions on here to make sure that I'm right and the book is wrong.

The title of the book is Physics 11, publisher is McGraw-Hill Ryerson
 
Thank you for your reply. This kind of information is very helpful for us here at PF.

Cheers!
 
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