## 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.

1. anonymous

Here's an illustration of te|dw:1444290768592:dw|

2. anonymous

The FBD I drew: |dw:1444290900790:dw| where N = normal force, W = weight, and f_k = kinetic friction

3. anonymous

And finally, the kinetic diagram describing the motion according to Newton's 2nd Law: |dw:1444291042878:dw|

4. anonymous

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

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

lbs is the unit of force, so I'd guess yes.

7. anonymous

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

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

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

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

$a = \frac{ 8t ^{2} - 2.4}{ 0.373 } = 21.4t ^{2} - 6.43$ approximately.

12. ParthKohli

Fair enough. Now$\int \limits_{0}^{2} adt = \int \limits_{4}^{v}dv$

13. ParthKohli

That makes sense right? So$\int \limits_0^2 (21.4t^2 - 6.43)dt = v - 4$

14. anonymous

Okay, and that comes out to v = 48.2 ft/s (roughly). Trying now, hold please...

15. anonymous

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

Yeah, the problem is that acceleration isn't constant. It varies with time. :)

17. anonymous

Huh. I guess I should probably just assume that acceleration is never constant unless the problem explicitly states otherwise...

18. anonymous

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!