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Homework Help: Work and Friction Question

  1. Sep 25, 2010 #1
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    I'm having trouble with numer 3. So work by friction is equal to delta K, which is equal to .5mv^2. So I used the kinematics equation v^2=u^2 + 2ad, where u^2 is the velocity found in number 2 squared. Then I plug in this value of v^2 into the kinetic energy equation to find work, but its not coming up as the right answer. Any ideas?

    Maybe I'm using the wrong value for acceleration? I've tried many methods for finding a, such as a=u[k]N or a=T/m, but none seem to work.
     
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  3. Sep 25, 2010 #2

    PhanthomJay

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    no, that is not correct. Net work done is delta K, which includes the work done by all forces.
    To find the acceleration. you have to draw a free body diagram of all forces acting. But you don't need to find the acceleration. Just find the friction force, then W_friction = dot product of friction force times displacemnt.
     
  4. Sep 25, 2010 #3
    Ok so about that free body diagram. This is a weird angle situation. So I have that upper angle as 33, so should the actual angle to the horizontal for finding mg be 90-33=57? So would Normal force be mgcos57? Or am I doing this wrong
     
  5. Sep 26, 2010 #4

    PhanthomJay

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    The gravity force, mg, acting on the block, always points vertically down, toward earth's center, regardless of whether the block is on a flat or sloped surface. It's the normal force that changes when on the incline. Since there is no movement of the block on the incline in the direction perpendicular to the incline, the normal force acting on it , perpendicular to the incline, must be equal and opposite to the algebraic sum of the components of all other forces perpendicular to the incline, per Newton 1. In this case, since friction and tension forces are parallel to the incline, then only the gravity force is the other force which has a component perpendicular to the incline, and that component is not mg cos 57, it's (mg)(___??___).
     
  6. Sep 26, 2010 #5
    Well I've tried mgcos57, mgcos33, mgsin33, mgsin57, but none of those worked. Am I still missing it?
     
  7. Sep 26, 2010 #6

    PhanthomJay

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    mgcos33 is the same as mgsin57; mgcos57 is the same as mgsin33. So essentially, you have 2 choices....is the gravity component, perpendicular to the incline, mgcoscos33 or mgcos57? Now since I've already noted that mgcos57 is wrong, then by elimination, it's mgcos 33, and the the normal force must be mgcos33 acting toward the block perpendicular to the incline, per Newton 1.. You should prove this to yourself by using geometry and trig....it is sometimes not readily apparent. Now calculate the friction force, and then, the work done by it.
     
  8. Sep 26, 2010 #7
    Ok I'm still not getting the right answer. Work is force * displacement, so the friction force is u[k]N, which we said to be .3*mgcos33 *1.6 m, which gives 51.33, but thats not the right answer.
     
  9. Sep 26, 2010 #8

    PhanthomJay

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    Call its magnitude 51.3 N, or even 50 N , using significant figures, but what is the direction of the friction force and what is the direction of its movement? Remember Work is the dot product of Force and Dispalcement.
     
  10. Sep 27, 2010 #9
    Ah ok I see this should be negative. Thanks!
     
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