1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Block Sliding up an Incline

  1. Jun 22, 2015 #1

    Drakkith

    User Avatar

    Staff: Mentor

    1. The problem statement, all variables and given/known data
    A mass m = 16 kg is pulled along a horizontal floor, with a coefficient of kinetic friction μk = 0.09, for a distance d = 7.6 m. Then the mass is continued to be pulled up a frictionless incline that makes an angle θ = 34° with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 34° (thus on the incline it is parallel to the surface) and has a tension T = 46 N.

    blockonincline.png

    1)
    What is the work done by tension before the block gets to the incline?
    2)
    What is the work done by friction as the block slides on the flat horizontal surface?
    3)
    What is the speed of the block right before it begins to travel up the incline?
    4)
    How far up the incline does the block travel before coming to rest?
    5)
    What is the work done by gravity as it comes to rest?


    2. Relevant equations


    3. The attempt at a solution
    First and foremost, I don't need help with every question. I've already solved 1-3. My problem is that I feel I'm just doing a 'trial and error' approach. I tend to get most of these questions wrong the first time and even after getting them right I don't usually feel that I've learned anything. For example, I just did a set of problems nearly identical to this one except that the horizontal plane was frictionless. When I look back on how I solved for the distance the block traveled up the incline, I don't even know why I solved it that way or why it even works.

    Specifically, I found the kinetic energy of the block right before it hit the ramp, 478 J, and used the following equation: 478 + Wt + Wg +Wf = 0
    Where Wt = work done by tension
    Wg = work done by gravity
    Wf = work done by friction

    I don't even know why I used this equation. On the very next problem (the one above) I found myself wondering why the kinetic energy (200 J) was on the left side and questioning which side it should be on and whether it should be negative or positive. In short, I feel like I have absolutely no idea what's going on and I don't know how to fix this problem.
     
  2. jcsd
  3. Jun 22, 2015 #2

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Did you get KE = 478 or 200 at the bottom of the incline? I get 200.

    The net work will equal the change in kinetic energy. Are we assuming the block starts from rest?

    You’ve got work due to tension, and negative work due to friction. For the work due to tension, only consider the component in the direction of motion (W = T d cos34).

    For friction be careful of your normal force. It’s not simply weight. You have to subtract the vertical component of tension.
     
  4. Jun 22, 2015 #3

    Drakkith

    User Avatar

    Staff: Mentor

    Apparently I can't read. I didn't see that the incline plane is frictionless. So I got the answer for Q4, 4.79 meters.

    Still, doing physics homework is a borderline nightmare for me. I feel utterly lost most of the time, and then I make it a thousand times harder on myself by making a simple mistake like this. I did the same thing last night where I had the sign of a force wrong and spent an hour losing my mind before finding the error.

    It's 200J. 478 was for a previous problem where the flat plane is frictionless and the ramp has friction.

    Let me ask you this: When bringing an object to a stop, does the total work just equal the initial kinetic energy, or is it the negative of the initial kinetic energy?
    For example, in this problem if I set Wtotal equal to -200 then:
    -200 = Wt - Wg
    -200 = FtX - FgX
    -200 = 46X - sin34(16)(9.81)X
    -200 = X(46-87.77)
    X = -200/-41.77
    X = 4.79

    So the work done has to be negative because it is done against the direction of motion, right? This didn't even click for me until just now.
     
  5. Jun 22, 2015 #4

    tony873004

    User Avatar
    Science Advisor
    Gold Member

    Total work equals the change in KE. So if you start with +200 J of KE, and you end with 0, then your change in KE = -200, so your work must be -200 as well.
    This is almost exactly like the example in your signature line. If you've got $200 in your bank account and you want to end with $0 in your account, you need to spend (negative, or red to an accountant) $200.
     
  6. Jun 22, 2015 #5
    Are you studying the conservation of energy? What components constitute the energy sum?
     
  7. Jun 22, 2015 #6

    Drakkith

    User Avatar

    Staff: Mentor

    Okay, that makes sense. Thanks Tony.

    Yes.

    Are you referring to U+K?
     
  8. Jun 23, 2015 #7
    Yes. What are the kinetic and potential energies when the block comes to rest? What is the only thing you don't know about the block at this point?

    Edit: I just saw that you solved part 4 above. Sorry.
     
    Last edited: Jun 23, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Block Sliding up an Incline
Loading...