rachel_and
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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rachel_and
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|dw:1351051208235:dw|
3psilon
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I'd start off with a free body diagram and sum of the forces
3psilon
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So do you know the x and y component of the force? Do you know how to get it?
rachel_and
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i know that there is friction, gravity, and the Fsintheta. do you add all three together?
rachel_and
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or subtract Fsintheta-(friction-gravity)
rachel_and
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friction+gravity
Yahoo!
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@rachel_and u knw to Draw FBD
3psilon
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|dw:1351052476768:dw|
3psilon
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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
Yahoo!
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|dw:1351052525480:dw|
3psilon
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\[Fsin \theta + Fn = 0 \]
\[Fcos \theta + Ff(friction) - mg = ma \]
Manipulate those equations to solve for force
Yahoo!
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@rachel_and it shuld be Fsintheta-(friction+gravity)
Yahoo!
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@3psilon Fcosx - N = 0
Yahoo!
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Normal Reaction is a Contact Force...
3psilon
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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
rachel_and
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So does F equal...
F(sin62-.497cos62) +2.48*9.8=0
3psilon
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|dw:1351053539218:dw|
Algebraic!
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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.
3psilon
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Oh constant speed so no acceleration ! Sorry :P
rachel_and
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I have the answer as 57.62 but apparently that is incorrect and I typed what @Algebraic! had
3psilon
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is your calc in degrees ?
rachel_and
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Yes
Algebraic!
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so did you get it right?
3psilon
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make sure you multiply by the distance
rachel_and
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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
Algebraic!
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yep. make sure it's not a sig. fig. or units error. pretty sure this answer is correct.
Algebraic!
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make sure you're supposed to be using 9.8 for g also.
rachel_and
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No sig figs or units here. There's no chance that work is negative, right?
Algebraic!
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no F and displacement are both upwards.
rachel_and
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sin62- .497cos62= .6496
2.48*9.8/ans=37.413
ans*1.54= 57.62
Algebraic!
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yep.
Algebraic!
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web assign?
Algebraic!
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mastering physics?
rachel_and
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web assignment!
Algebraic!
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any chance you could screen shot the page and post it?
rachel_and
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3psilon
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Is it the angle?
3psilon
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Should it be 28 instead ? 90 - 62?
Locust_Grub
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3psilon
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Try 28 @rachel_and since it is 28 degrees with respect to the vertical
Locust_Grub
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Locust grub needs help with his question too >:L
Algebraic!
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hmm not sure @rachel_and ...
Locust_Grub
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D:<
rachel_and
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Assignment deadline has past so I'll wait until the answers and solutions are posted. Thanks ya'll!!
Algebraic!
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don't forget to post back here with the right solution! :)
rachel_and
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Will do!
Algebraic!
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ty!