## anonymous 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. anonymous

|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. anonymous

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

5. anonymous

or subtract Fsintheta-(friction-gravity)

6. anonymous

friction+gravity

7. anonymous

@rachel_and u knw to Draw FBD

8. 3psilon

|dw:1351052476768:dw|

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. anonymous

|dw:1351052525480:dw|

11. 3psilon

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

12. anonymous

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

13. anonymous

@3psilon Fcosx - N = 0

14. anonymous

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. anonymous

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

17. 3psilon

|dw:1351053539218:dw|

18. 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.

19. 3psilon

Oh constant speed so no acceleration ! Sorry :P

20. anonymous

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. anonymous

Yes

23. anonymous

so did you get it right?

24. 3psilon

make sure you multiply by the distance

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. anonymous

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

29. anonymous

no F and displacement are both upwards.

30. anonymous

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

31. anonymous

yep.

32. anonymous

web assign?

33. anonymous

mastering physics?

34. anonymous

web assignment!

35. anonymous

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

36. anonymous

37. 3psilon

Is it the angle?

38. 3psilon

Should it be 28 instead ? 90 - 62?

39. anonymous

40. 3psilon

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

41. anonymous

Locust grub needs help with his question too >:L

42. anonymous

hmm not sure @rachel_and ...

43. anonymous

D:<

44. anonymous

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

45. anonymous

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

46. anonymous

Will do!

47. anonymous

ty!