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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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rachel_andBest ResponseYou've already chosen the best response.0
dw:1351051208235:dw
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
I'd start off with a free body diagram and sum of the forces
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
So do you know the x and y component of the force? Do you know how to get it?
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
i know that there is friction, gravity, and the Fsintheta. do you add all three together?
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
or subtract Fsintheta(frictiongravity)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@rachel_and u knw to Draw FBD
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
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
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
\[Fsin \theta + Fn = 0 \] \[Fcos \theta + Ff(friction)  mg = ma \] Manipulate those equations to solve for force
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
@rachel_and it shuld be Fsintheta(friction+gravity)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.0
Normal Reaction is a Contact Force...
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
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
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
So does F equal... F(sin62.497cos62) +2.48*9.8=0
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
'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.
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
Oh constant speed so no acceleration ! Sorry :P
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
I have the answer as 57.62 but apparently that is incorrect and I typed what @Algebraic! had
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
is your calc in degrees ?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
so did you get it right?
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
make sure you multiply by the distance
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
No, 57.62 is incorrect. If you plug in the values, do you get the same answer? And I multiplied by the distance
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
yep. make sure it's not a sig. fig. or units error. pretty sure this answer is correct.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
make sure you're supposed to be using 9.8 for g also.
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
No sig figs or units here. There's no chance that work is negative, right?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
no F and displacement are both upwards.
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
sin62 .497cos62= .6496 2.48*9.8/ans=37.413 ans*1.54= 57.62
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
any chance you could screen shot the page and post it?
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
Should it be 28 instead ? 90  62?
 one year ago

3psilonBest ResponseYou've already chosen the best response.1
Try 28 @rachel_and since it is 28 degrees with respect to the vertical
 one year ago

Locust_GrubBest ResponseYou've already chosen the best response.0
Locust grub needs help with his question too >:L
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
hmm not sure @rachel_and ...
 one year ago

rachel_andBest ResponseYou've already chosen the best response.0
Assignment deadline has past so I'll wait until the answers and solutions are posted. Thanks ya'll!!
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.2
don't forget to post back here with the right solution! :)
 one year ago
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