anonymous
  • anonymous
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.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1351051208235:dw|
3psilon
  • 3psilon
I'd start off with a free body diagram and sum of the forces
3psilon
  • 3psilon
So do you know the x and y component of the force? Do you know how to get it?

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More answers

anonymous
  • anonymous
i know that there is friction, gravity, and the Fsintheta. do you add all three together?
anonymous
  • anonymous
or subtract Fsintheta-(friction-gravity)
anonymous
  • anonymous
friction+gravity
anonymous
  • anonymous
@rachel_and u knw to Draw FBD
3psilon
  • 3psilon
|dw:1351052476768:dw|
3psilon
  • 3psilon
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
anonymous
  • anonymous
|dw:1351052525480:dw|
3psilon
  • 3psilon
\[Fsin \theta + Fn = 0 \] \[Fcos \theta + Ff(friction) - mg = ma \] Manipulate those equations to solve for force
anonymous
  • anonymous
@rachel_and it shuld be Fsintheta-(friction+gravity)
anonymous
  • anonymous
@3psilon Fcosx - N = 0
anonymous
  • anonymous
Normal Reaction is a Contact Force...
3psilon
  • 3psilon
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
anonymous
  • anonymous
So does F equal... F(sin62-.497cos62) +2.48*9.8=0
3psilon
  • 3psilon
|dw:1351053539218:dw|
anonymous
  • anonymous
'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.
3psilon
  • 3psilon
Oh constant speed so no acceleration ! Sorry :P
anonymous
  • anonymous
I have the answer as 57.62 but apparently that is incorrect and I typed what @Algebraic! had
3psilon
  • 3psilon
is your calc in degrees ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
so did you get it right?
3psilon
  • 3psilon
make sure you multiply by the distance
anonymous
  • anonymous
No, 57.62 is incorrect. If you plug in the values, do you get the same answer? And I multiplied by the distance
anonymous
  • anonymous
yep. make sure it's not a sig. fig. or units error. pretty sure this answer is correct.
anonymous
  • anonymous
make sure you're supposed to be using 9.8 for g also.
anonymous
  • anonymous
No sig figs or units here. There's no chance that work is negative, right?
anonymous
  • anonymous
no F and displacement are both upwards.
anonymous
  • anonymous
sin62- .497cos62= .6496 2.48*9.8/ans=37.413 ans*1.54= 57.62
anonymous
  • anonymous
yep.
anonymous
  • anonymous
web assign?
anonymous
  • anonymous
mastering physics?
anonymous
  • anonymous
web assignment!
anonymous
  • anonymous
any chance you could screen shot the page and post it?
anonymous
  • anonymous
1 Attachment
3psilon
  • 3psilon
Is it the angle?
3psilon
  • 3psilon
Should it be 28 instead ? 90 - 62?
anonymous
  • anonymous
1 Attachment
3psilon
  • 3psilon
Try 28 @rachel_and since it is 28 degrees with respect to the vertical
anonymous
  • anonymous
Locust grub needs help with his question too >:L
anonymous
  • anonymous
hmm not sure @rachel_and ...
anonymous
  • anonymous
D:<
anonymous
  • anonymous
Assignment deadline has past so I'll wait until the answers and solutions are posted. Thanks ya'll!!
anonymous
  • anonymous
don't forget to post back here with the right solution! :)
anonymous
  • anonymous
Will do!
anonymous
  • anonymous
ty!

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