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Finding Average Force Applied On A Slope

  1. Aug 8, 2013 #1
    Hi all. I am marking GCSE paper that I did myself. There is 1 problem that I do not seems to understand. Please help me. Thank you very much.

    1. The problem statement, all variables and given/known data
    Daniel pulls a 3kg box up a slope from rest as shown (Attached)
    Given that the average friction of a slope is 2N and the speed at which the box reaches the top is 4m/s. What is the average form ce exerted by Daniel to pull up the box from bottom to top?

    3. The attempt at a solution
    I am stuck at this part because the question does not indicate the time taken nor the acceleration. I am not able to use F=MA to find. I only have the acceleration acting on the box due to gravity is approximately 10m/s2. I am not able to use the speed to my advantage in this question.

    Weight = m x g
    = 3 x 10
    = 30 N

    I am not sure what is my next step. Hope someone can help me. Thank you!
     

    Attached Files:

  2. jcsd
  3. Aug 8, 2013 #2
    Perhaps you could use the work-energy approach?
     
  4. Aug 8, 2013 #3

    gneill

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    Staff: Mentor

    Hi mynameiscosine, Welcome to Physics Forums.

    Actually, you can make use of the given velocity; Can you think of a kinematic equation which relates velocity, acceleration, and distance?
     
  5. Aug 8, 2013 #4

    Zondrina

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    Homework Helper

    Hint : Remember that the initial velocity of Daniel is 0.
     
  6. Aug 8, 2013 #5
    I can do this but I am not sure if I am over-complicating this.

    a = acceleration
    v = velocity

    a = (4-0)/t
    = 4/t

    integrate a = v
    v = 4 ln (t)

    When v = 4
    ln (t) = 1
    t = e^1

    therefore,
    a = 4/e^1
    = 1.4715... m/s^2

    Force = m x a
    = 3 x 1.4715
    = 4.4145... N

    I am not sure if my next one is right,

    E = 30 N x 12
    = 360J

    Work done = R x 15
    R = 360/15
    = 24N

    What can I do from here?

    Thank you!
     
  7. Aug 8, 2013 #6
    This is not correct. In the equation a = (4 - 0)/t, t denotes the total time. This is not the "current" time you can use to go from acceleration to velocity via integration.

    This attempt is better, but still not entirely correct. You assign only potential energy to E, while it must be total energy.
     
  8. Aug 8, 2013 #7
    Oh I see. Can I ask you the reason why I cannot go from acceleration to velocity via integration? I actually thought that t is the same for both equations? Thank you.

    Which energy have I not considered into total energy?

    My initial thought was,
    Law of conservation of energy,
    so,
    EPotential = EKinetic
    mass x acceleration due to gravity x height = 0.5 x mass x (velocity)2
    Since 4m/s is not the average speed, I cannot use 4.
    Am I wrong?

    What should I do next?

    Thank you!
     
  9. Aug 8, 2013 #8

    gneill

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    Staff: Mentor

    You could go from acceleration to velocity via integration if you were given the time involved. But you are not given the time. It's usually not helpful to introduce new variables for which you are not given values.

    There is another kinematic equation that relates velocity, acceleration, and distance without using any other variables. Check your notes or text.
    What happened to the energy lost due to friction?

    EDIT:
    Consider what work the applied force has to do. It has to:

    1) Raise the box by some amount against gravity
    2) Provide the box with some final velocity
    3) Create some heat due to sliding friction along the ramp.

    Can you identify the types of energy associated with each of the above and then write an equation to balance the work done?
     
    Last edited: Aug 8, 2013
  10. Aug 8, 2013 #9
    The other equation is
    d = vt - 0.5at2
    v is the initial velocity.

    But I cannot figure out how this can help me find acceleration as time is not given to me.

    I am sorry, how do I calculate the energy lost due to friction in this case?
    Thank you!
     
  11. Aug 8, 2013 #10

    gneill

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    Staff: Mentor

    Again, this equation introduces the variable t (and it's just a generalized variant of the one you tried previously). No, the equation you're looking for is: vf2 - vi2 = 2ad. Note that only velocity, acceleration, and distance variables are used in this equation.
    What's the frictional force? What's the distance over which it acts?
     
  12. Aug 8, 2013 #11
    Oh I got what you mean now! Sorry for getting it slow!

    Correct me if I am wrong:

    Energy loss due to friction = F x d
    2N x 15 = 30J

    Epotential + Ekinetic + Efrictional
    = (30 x 12) + (0.5 x 3 x 42) + 30
    = 360 + 24 + 30
    = 414J

    Am I understanding this part correctly?

    After this, I must have understanding something wrongly,

    Work done = R x 15
    R = 414/15
    R = 27.6N

    Resultant force = Applied force - Frictional Force
    27.6 = x - 2
    x = 29.6N

    Where is my mistake?

    Thank you!
     
  13. Aug 8, 2013 #12

    gneill

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    Staff: Mentor

    Yes, so far so good :smile:
    There's no need to subtract the frictional force --- it's a component of the work done by the applied force, so it's already accounted for. Your "R" is the applied force; It's the force applied to the box in order to accomplish the work specified, which includes the work done against friction.
     
  14. Aug 8, 2013 #13
    Oh wow. Interesting!
    You made me see my mistakes in understanding dynamics!

    I think sometimes I am too focus in looking at this problem in one angle. I kept trying to introduce the variable 't' into the problem instead.

    Thank you so much for taking the time to explain to me. I fully understand now!
     
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