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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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
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|dw:1351051208235:dw|
I'd start off with a free body diagram and sum of the forces
So do you know the x and y component of the force? Do you know how to get it?

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

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