What is the horizontal acceleration of a 7.00-kg block with two given forces?

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Homework Help Overview

The problem involves determining the horizontal acceleration of a 7.00-kg block subjected to two forces, F1 and F2, with specified magnitudes. The context includes the application of Newton's second law, where the net force acting on the block is to be calculated.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations of forces and the resulting acceleration, with some questioning the accuracy of their results. There is a focus on the potential omission of a coefficient of friction and the sensitivity of the homework program to significant figures.

Discussion Status

The discussion has evolved with participants sharing their attempts and frustrations regarding the correctness of their answers. Some have suggested that the issue may lie in the details of the homework setup, such as significant figures, while others have confirmed that adjusting their answer format resolved the issue.

Contextual Notes

Participants mention the possibility of missing information, such as a coefficient of friction, which could affect the calculations. There is also a concern about the homework program's sensitivity to significant figures.

KachinaEarth
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Homework Statement


Two forces, F1 and F2, act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are F1=65.9 N and F2=22.7 N. Take the positive direction to be to the right. Find the horizontal acceleration of the block, including sign.

http://img522.imageshack.us/my.php?image=forcett9.jpg


Homework Equations



[tex]\Sigma[/tex]F = ma or a = [tex]\Sigma[/tex]F / m

[tex]\Sigma[/tex]F - Net Force (F1 + F2 = Net F)
m - mass
a - acceleration


The Attempt at a Solution



F1 = 65.9N
F1 * Sin = 61.9N
F1 * Cos = 22.6N

F2 = 22.7 N
F2 Opposite Force = -22.7 N

(-) 22.7 N + (+) 22.54 N = 7 kg*a
(-).16 N = 7 kg*a
(-).16 N / 7 kg = a
(-)0.023 m/s[tex]^{2}[/tex]


I'm not quite sure what I'm doing wrong as when I try to use this answer it is marked as incorrect so any help would be most appreciated.
 
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-0.023 m/s^2 was marked as incorrect?
 
Correct; my answer is marked as incorrect but I'm not sure why as it seems to be the only answer I can arrive at using the formula taught to me both in my class and in my recitation lab.
 
KachinaEarth said:
Correct; my answer is marked as incorrect but I'm not sure why as it seems to be the only answer I can arrive at using the formula taught to me both in my class and in my recitation lab.

Did you post the question exactly as it is? Did you maybe mix up F1 and F2?
 
Directly as is. I was very careful to type everything word, and number, for word.

Is there perhaps an error with the homework you think?
 
KachinaEarth said:
Directly as is. I was very careful to type everything word, and number, for word.

Is there perhaps an error with the homework you think?

yeah I think so... I suspect maybe there's a coefficient of friction that's supposed to be in the problem... but they left out...

Unless the program is sensitive about significant figures or something?
Maybe -0.0230m/s^2 will work? Not sure...
 
I tried the extra zero at the end of -0.0230 m/s^2 and it took the answer, finally, so I think the program is just that sensitive. Thank you so much for your help. I was starting to stress so much I don't think I would have thought to keep adding on significant numbers.
 
KachinaEarth said:
I tried the extra zero at the end of -0.0230 m/s^2 and it took the answer, finally, so I think the program is just that sensitive. Thank you so much for your help. I was starting to stress so much I don't think I would have thought to keep adding on significant numbers.

No prob. Glad it went through. :smile:
 

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