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rachel_and

  • 3 years ago

A 2.48 kg block is pushed 1.54 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 62 degrees with the horizontal. The acceleration of gravity is 9.8 m/s^2. If the coefficient of kinetic friction between the block and wall is .497, find the work done by F.

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  1. rachel_and
    • 3 years ago
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    |dw:1351051208235:dw|

  2. 3psilon
    • 3 years ago
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    I'd start off with a free body diagram and sum of the forces

  3. 3psilon
    • 3 years ago
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    So do you know the x and y component of the force? Do you know how to get it?

  4. rachel_and
    • 3 years ago
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    i know that there is friction, gravity, and the Fsintheta. do you add all three together?

  5. rachel_and
    • 3 years ago
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    or subtract Fsintheta-(friction-gravity)

  6. rachel_and
    • 3 years ago
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    friction+gravity

  7. Yahoo!
    • 3 years ago
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    @rachel_and u knw to Draw FBD

  8. 3psilon
    • 3 years ago
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    |dw:1351052476768:dw|

  9. 3psilon
    • 3 years ago
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    We need force cause work = force * distance . So once we find force we can plug it into that equation and we have the distance already

  10. Yahoo!
    • 3 years ago
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    |dw:1351052525480:dw|

  11. 3psilon
    • 3 years ago
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    \[Fsin \theta + Fn = 0 \] \[Fcos \theta + Ff(friction) - mg = ma \] Manipulate those equations to solve for force

  12. Yahoo!
    • 3 years ago
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    @rachel_and it shuld be Fsintheta-(friction+gravity)

  13. Yahoo!
    • 3 years ago
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    @3psilon Fcosx - N = 0

  14. Yahoo!
    • 3 years ago
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    Normal Reaction is a Contact Force...

  15. 3psilon
    • 3 years ago
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    Oh yes sorry normal should be negative , and I wasn't getting your replies earlier @rachel_and I thought you were gone :P that's weird they show up now

  16. rachel_and
    • 3 years ago
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    So does F equal... F(sin62-.497cos62) +2.48*9.8=0

  17. 3psilon
    • 3 years ago
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    |dw:1351053539218:dw|

  18. Algebraic!
    • 3 years ago
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    'x' forces: \[N = F \cos 62\] 'y' forces: \[\mu _{k} N =\mu _{k} F \cos 62\] \[Fsin 62 = mg +\mu _{k} F \cos 62\] \[F(\sin 62- \mu _{k} \cos 62) = mg \] \[F= \frac{ mg }{ (\sin 62- \mu _{k} \cos 62)}\] calculate F; F*d = Work.

  19. 3psilon
    • 3 years ago
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    Oh constant speed so no acceleration ! Sorry :P

  20. rachel_and
    • 3 years ago
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    I have the answer as 57.62 but apparently that is incorrect and I typed what @Algebraic! had

  21. 3psilon
    • 3 years ago
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    is your calc in degrees ?

  22. rachel_and
    • 3 years ago
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    Yes

  23. Algebraic!
    • 3 years ago
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    so did you get it right?

  24. 3psilon
    • 3 years ago
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    make sure you multiply by the distance

  25. rachel_and
    • 3 years ago
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    No, 57.62 is incorrect. If you plug in the values, do you get the same answer? And I multiplied by the distance

  26. Algebraic!
    • 3 years ago
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    yep. make sure it's not a sig. fig. or units error. pretty sure this answer is correct.

  27. Algebraic!
    • 3 years ago
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    make sure you're supposed to be using 9.8 for g also.

  28. rachel_and
    • 3 years ago
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    No sig figs or units here. There's no chance that work is negative, right?

  29. Algebraic!
    • 3 years ago
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    no F and displacement are both upwards.

  30. rachel_and
    • 3 years ago
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    sin62- .497cos62= .6496 2.48*9.8/ans=37.413 ans*1.54= 57.62

  31. Algebraic!
    • 3 years ago
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    yep.

  32. Algebraic!
    • 3 years ago
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    web assign?

  33. Algebraic!
    • 3 years ago
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    mastering physics?

  34. rachel_and
    • 3 years ago
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    web assignment!

  35. Algebraic!
    • 3 years ago
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    any chance you could screen shot the page and post it?

  36. rachel_and
    • 3 years ago
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  37. 3psilon
    • 3 years ago
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    Is it the angle?

  38. 3psilon
    • 3 years ago
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    Should it be 28 instead ? 90 - 62?

  39. Locust_Grub
    • 3 years ago
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  40. 3psilon
    • 3 years ago
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    Try 28 @rachel_and since it is 28 degrees with respect to the vertical

  41. Locust_Grub
    • 3 years ago
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    Locust grub needs help with his question too >:L

  42. Algebraic!
    • 3 years ago
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    hmm not sure @rachel_and ...

  43. Locust_Grub
    • 3 years ago
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    D:<

  44. rachel_and
    • 3 years ago
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    Assignment deadline has past so I'll wait until the answers and solutions are posted. Thanks ya'll!!

  45. Algebraic!
    • 3 years ago
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    don't forget to post back here with the right solution! :)

  46. rachel_and
    • 3 years ago
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    Will do!

  47. Algebraic!
    • 3 years ago
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    ty!

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