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.

- anonymous

- katieb

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- anonymous

|dw:1351051208235:dw|

- 3psilon

I'd start off with a free body diagram and sum of the forces

- 3psilon

So do you know the x and y component of the force? Do you know how to get it?

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## More answers

- anonymous

i know that there is friction, gravity, and the Fsintheta. do you add all three together?

- anonymous

or subtract Fsintheta-(friction-gravity)

- anonymous

friction+gravity

- anonymous

@rachel_and u knw to Draw FBD

- 3psilon

|dw:1351052476768:dw|

- 3psilon

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

- anonymous

|dw:1351052525480:dw|

- 3psilon

\[Fsin \theta + Fn = 0 \]
\[Fcos \theta + Ff(friction) - mg = ma \]
Manipulate those equations to solve for force

- anonymous

@rachel_and it shuld be Fsintheta-(friction+gravity)

- anonymous

@3psilon Fcosx - N = 0

- anonymous

Normal Reaction is a Contact Force...

- 3psilon

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

- anonymous

So does F equal...
F(sin62-.497cos62) +2.48*9.8=0

- 3psilon

|dw:1351053539218:dw|

- anonymous

'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

Oh constant speed so no acceleration ! Sorry :P

- anonymous

I have the answer as 57.62 but apparently that is incorrect and I typed what @Algebraic! had

- 3psilon

is your calc in degrees ?

- anonymous

Yes

- anonymous

so did you get it right?

- 3psilon

make sure you multiply by the distance

- anonymous

No, 57.62 is incorrect. If you plug in the values, do you get the same answer?
And I multiplied by the distance

- anonymous

yep. make sure it's not a sig. fig. or units error. pretty sure this answer is correct.

- anonymous

make sure you're supposed to be using 9.8 for g also.

- anonymous

No sig figs or units here. There's no chance that work is negative, right?

- anonymous

no F and displacement are both upwards.

- anonymous

sin62- .497cos62= .6496
2.48*9.8/ans=37.413
ans*1.54= 57.62

- anonymous

yep.

- anonymous

web assign?

- anonymous

mastering physics?

- anonymous

web assignment!

- anonymous

any chance you could screen shot the page and post it?

- anonymous

##### 1 Attachment

- 3psilon

Is it the angle?

- 3psilon

Should it be 28 instead ? 90 - 62?

- anonymous

##### 1 Attachment

- 3psilon

Try 28 @rachel_and since it is 28 degrees with respect to the vertical

- anonymous

Locust grub needs help with his question too >:L

- anonymous

hmm not sure @rachel_and ...

- anonymous

D:<

- anonymous

Assignment deadline has past so I'll wait until the answers and solutions are posted. Thanks ya'll!!

- anonymous

don't forget to post back here with the right solution! :)

- anonymous

Will do!

- anonymous

ty!

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