## 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. one year ago one year ago

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1. rachel_and Group Title

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2. 3psilon Group Title

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

3. 3psilon Group Title

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

4. rachel_and Group Title

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

5. rachel_and Group Title

or subtract Fsintheta-(friction-gravity)

6. rachel_and Group Title

friction+gravity

7. Yahoo! Group Title

@rachel_and u knw to Draw FBD

8. 3psilon Group Title

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9. 3psilon Group Title

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! Group Title

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11. 3psilon Group Title

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

12. Yahoo! Group Title

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

13. Yahoo! Group Title

@3psilon Fcosx - N = 0

14. Yahoo! Group Title

Normal Reaction is a Contact Force...

15. 3psilon Group Title

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 Group Title

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

17. 3psilon Group Title

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18. Algebraic! Group Title

'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 Group Title

Oh constant speed so no acceleration ! Sorry :P

20. rachel_and Group Title

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

21. 3psilon Group Title

is your calc in degrees ?

22. rachel_and Group Title

Yes

23. Algebraic! Group Title

so did you get it right?

24. 3psilon Group Title

make sure you multiply by the distance

25. rachel_and Group Title

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! Group Title

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

27. Algebraic! Group Title

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

28. rachel_and Group Title

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

29. Algebraic! Group Title

no F and displacement are both upwards.

30. rachel_and Group Title

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

31. Algebraic! Group Title

yep.

32. Algebraic! Group Title

web assign?

33. Algebraic! Group Title

mastering physics?

34. rachel_and Group Title

web assignment!

35. Algebraic! Group Title

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

36. rachel_and Group Title

37. 3psilon Group Title

Is it the angle?

38. 3psilon Group Title

Should it be 28 instead ? 90 - 62?

39. Locust_Grub Group Title

40. 3psilon Group Title

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

41. Locust_Grub Group Title

Locust grub needs help with his question too >:L

42. Algebraic! Group Title

hmm not sure @rachel_and ...

43. Locust_Grub Group Title

D:<

44. rachel_and Group Title

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

45. Algebraic! Group Title

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

46. rachel_and Group Title

Will do!

47. Algebraic! Group Title

ty!