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anonymous

  • one year ago

Heya, anyone who's good at Dynamics. I have a question involving kinetic friction. There's a box weighing 12 lb moving at a velocity of 4 ft/s on the ground when force F = (8t^2) lb is applied. Coefficient of kinetic friction is 0.2. It wants me to find the velocity at time equals 2 seconds. I've gone through it a couple of times and my answer keeps coming up as v = 163 ft/s. This is apparently wrong though, and I'm running out of attempts. Can someone help me? I'd be more than happy to tell you my methodology for the answer I've got.

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  1. anonymous
    • one year ago
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    Here's an illustration of te|dw:1444290768592:dw|

  2. anonymous
    • one year ago
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    The FBD I drew: |dw:1444290900790:dw| where N = normal force, W = weight, and f_k = kinetic friction

  3. anonymous
    • one year ago
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    And finally, the kinetic diagram describing the motion according to Newton's 2nd Law: |dw:1444291042878:dw|

  4. anonymous
    • one year ago
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    I solved for mass using weight, which got me approx. 0.373 slugs, I solved for forces of y to determine that magnitude of N = W, which is 12 lb. I plugged that into the equation for kinetic friction, which got me f_k = 2.4 lb. Summed forces of x equal to the motion to find acceleration at t = 2s, which got me 79.4 ft/s^2. I then used the constant acceleration equation where v_0 = 4 ft/s, a = 79.4 ft/s^2, and t = 2s. This calculation comes out to v = 163 ft/s. Mastering Engineering tells me that this is the wrong answer. But I don't know what I'm doing wrong.

  5. ParthKohli
    • one year ago
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    I'm not too good with imperial units, so I need one clarification: 12 lbs means that the box weighs 12 lbs right? Like we don't have to multiply that by the acceleration due to gravity, eh?

  6. ParthKohli
    • one year ago
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    lbs is the unit of force, so I'd guess yes.

  7. anonymous
    • one year ago
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    I believe that's the weight, yeah. But to be fair, the problem isn't too clear on that. I figured it was weight, making slug the unit of mass.

  8. ParthKohli
    • one year ago
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    In that case, we have two forces acting on the block: 1. \(8t^2\) lbs forward 2. \(0.2 \times 12 = 2.4\) lbs backward. Now\[\int F dt =\int m dv \]or\[\int a dt =\int dv \]Both equations are the same (the second equation is the first divided by mass on both sides). Did you use these?

  9. anonymous
    • one year ago
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    It didn't strike me as terribly relevant, no. I figured since \[\Sigma F _{x} = F - f _{k} = ma\], and I could find mass using the weight, that I could solve for acceleration that way. Should I not have done that, and used the integration method instead?

  10. ParthKohli
    • one year ago
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    Yes, alright, so you solved for the acceleration and it should turn out to be a function of \(t\). What did you get?

  11. anonymous
    • one year ago
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    \[a = \frac{ 8t ^{2} - 2.4}{ 0.373 } = 21.4t ^{2} - 6.43\] approximately.

  12. ParthKohli
    • one year ago
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    Fair enough. Now\[\int \limits_{0}^{2} adt = \int \limits_{4}^{v}dv\]

  13. ParthKohli
    • one year ago
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    That makes sense right? So\[\int \limits_0^2 (21.4t^2 - 6.43)dt = v - 4\]

  14. anonymous
    • one year ago
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    Okay, and that comes out to v = 48.2 ft/s (roughly). Trying now, hold please...

  15. anonymous
    • one year ago
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    Okay, yeah, that worked. Huh. Why wouldn't the other kinematics equation I tried work? I thought for sure that acceleration was constant...

  16. ParthKohli
    • one year ago
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    Yeah, the problem is that acceleration isn't constant. It varies with time. :)

  17. anonymous
    • one year ago
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    Huh. I guess I should probably just assume that acceleration is never constant unless the problem explicitly states otherwise...

  18. anonymous
    • one year ago
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    That said, I appreciate your help! I gave you a medal. I'm going to go ahead and close this thread now if it's alright with you. Thanks again!

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