anonymous
  • anonymous
Please I need help with a new question, this has a solution that I can't still dig. See the attached JPG
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
This is the solution I cant dig. Any help is welcome.
1 Attachment
Mani_Jha
  • Mani_Jha
The equations for motion under constant acceleration are: \[v=u+at\] \[s=ut+at ^{2}/2\] In this case u=0. And the question acts for motion ALONG the board. But the given acceleration is vertical. You've to break it into its components, such that one component is along the board. |dw:1333081367432:dw| So the acceleration along the board is a0sintheta. Now just substitute: \[a=a0\sin \theta\]
anonymous
  • anonymous
@ Mani_Jha :thnks for your reply, what about Weight?, ┬┐given that block's size is negligible we don't take it in account?

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anonymous
  • anonymous
Your approach is very much simpler than the one stated on the JPG.
anonymous
  • anonymous
@Mani_Jha :Just one more thing, the direction that you stated for ao/sin/theta is inwards, but the direction stated at the jpg is outwards (from A to C)!!!!.
anonymous
  • anonymous
Maybe I closed too early this question!
mos1635
  • mos1635
|dw:1333180717169:dw|

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