## rachel_and 3 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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1. rachel_and

|dw:1351051208235:dw|

2. 3psilon

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

3. 3psilon

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

4. rachel_and

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

5. rachel_and

or subtract Fsintheta-(friction-gravity)

6. rachel_and

friction+gravity

7. Yahoo!

@rachel_and u knw to Draw FBD

8. 3psilon

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9. 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

10. Yahoo!

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11. 3psilon

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

12. Yahoo!

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

13. Yahoo!

@3psilon Fcosx - N = 0

14. Yahoo!

Normal Reaction is a Contact Force...

15. 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

16. rachel_and

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

17. 3psilon

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18. Algebraic!

'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.

19. 3psilon

Oh constant speed so no acceleration ! Sorry :P

20. rachel_and

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

21. 3psilon

is your calc in degrees ?

22. rachel_and

Yes

23. Algebraic!

so did you get it right?

24. 3psilon

make sure you multiply by the distance

25. rachel_and

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

26. Algebraic!

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

27. Algebraic!

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

28. rachel_and

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

29. Algebraic!

no F and displacement are both upwards.

30. rachel_and

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

31. Algebraic!

yep.

32. Algebraic!

web assign?

33. Algebraic!

mastering physics?

34. rachel_and

web assignment!

35. Algebraic!

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

36. rachel_and

37. 3psilon

Is it the angle?

38. 3psilon

Should it be 28 instead ? 90 - 62?

39. Locust_Grub

40. 3psilon

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

41. Locust_Grub

Locust grub needs help with his question too >:L

42. Algebraic!

hmm not sure @rachel_and ...

43. Locust_Grub

D:<

44. rachel_and

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

45. Algebraic!

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

46. rachel_and

Will do!

47. Algebraic!

ty!