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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.
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Here's an illustration of tedw:1444290768592:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The FBD I drew: dw:1444290900790:dw where N = normal force, W = weight, and f_k = kinetic friction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And finally, the kinetic diagram describing the motion according to Newton's 2nd Law: dw:1444291042878:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I'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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2lbs is the unit of force, so I'd guess yes.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2In 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, alright, so you solved for the acceleration and it should turn out to be a function of \(t\). What did you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a = \frac{ 8t ^{2}  2.4}{ 0.373 } = 21.4t ^{2}  6.43\] approximately.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Fair enough. Now\[\int \limits_{0}^{2} adt = \int \limits_{4}^{v}dv\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2That makes sense right? So\[\int \limits_0^2 (21.4t^2  6.43)dt = v  4\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, and that comes out to v = 48.2 ft/s (roughly). Trying now, hold please...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, yeah, that worked. Huh. Why wouldn't the other kinematics equation I tried work? I thought for sure that acceleration was constant...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, the problem is that acceleration isn't constant. It varies with time. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Huh. I guess I should probably just assume that acceleration is never constant unless the problem explicitly states otherwise...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That 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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