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rachel_and
Group Title
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.
 2 years ago
 2 years ago
rachel_and Group Title
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.
 2 years ago
 2 years ago

This Question is Open

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
dw:1351051208235:dw
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
I'd start off with a free body diagram and sum of the forces
 2 years ago

3psilon Group TitleBest 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?
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
i know that there is friction, gravity, and the Fsintheta. do you add all three together?
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
or subtract Fsintheta(frictiongravity)
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
friction+gravity
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@rachel_and u knw to Draw FBD
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
dw:1351052476768:dw
 2 years ago

3psilon Group TitleBest 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
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
dw:1351052525480:dw
 2 years ago

3psilon Group TitleBest 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
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@rachel_and it shuld be Fsintheta(friction+gravity)
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
@3psilon Fcosx  N = 0
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Normal Reaction is a Contact Force...
 2 years ago

3psilon Group TitleBest 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
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
So does F equal... F(sin62.497cos62) +2.48*9.8=0
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
dw:1351053539218:dw
 2 years ago

Algebraic! Group TitleBest 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.
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
Oh constant speed so no acceleration ! Sorry :P
 2 years ago

rachel_and Group TitleBest 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
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
is your calc in degrees ?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
so did you get it right?
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
make sure you multiply by the distance
 2 years ago

rachel_and Group TitleBest 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
 2 years ago

Algebraic! Group TitleBest 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.
 2 years ago

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

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
No sig figs or units here. There's no chance that work is negative, right?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
no F and displacement are both upwards.
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
sin62 .497cos62= .6496 2.48*9.8/ans=37.413 ans*1.54= 57.62
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
web assign?
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
mastering physics?
 2 years ago

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
web assignment!
 2 years ago

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

3psilon Group TitleBest ResponseYou've already chosen the best response.1
Is it the angle?
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
Should it be 28 instead ? 90  62?
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.1
Try 28 @rachel_and since it is 28 degrees with respect to the vertical
 2 years ago

Locust_Grub Group TitleBest ResponseYou've already chosen the best response.0
Locust grub needs help with his question too >:L
 2 years ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.2
hmm not sure @rachel_and ...
 2 years ago

rachel_and Group TitleBest 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!!
 2 years ago

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

rachel_and Group TitleBest ResponseYou've already chosen the best response.0
Will do!
 2 years ago
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