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 2 years 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.
 2 years 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.

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

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

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

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

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

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

3psilon
 2 years ago
Best ResponseYou've already chosen the best response.1We need force cause work = force * distance . So once we find force we can plug it into that equation and we have the distance already

3psilon
 2 years ago
Best 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

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

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

3psilon
 2 years ago
Best ResponseYou've already chosen the best response.1Oh 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
 2 years ago
Best ResponseYou've already chosen the best response.0So does F equal... F(sin62.497cos62) +2.48*9.8=0

Algebraic!
 2 years ago
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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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