Did I solve this inclined plane problem correctly?

AI Thread Summary
The discussion revolves around solving a physics problem involving a block on an inclined plane with an initial velocity and kinetic friction. The original poster calculated the distance the block travels before stopping as 0.302 meters, using Newton's second law and kinematic equations. However, a mistake was identified in the application of the kinematic formula, specifically overlooking the need to square the initial velocity. This error led to a revised distance of 1.5 meters after correcting the calculation. The conversation highlights the importance of careful application of formulas in physics problems.
QuarkCharmer
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Homework Statement


I took a test today and I simply have to know if I got this question right! Otherwise it will haunt me for the whole weekend.

vil3tk.jpg

A block is at the bottom of an inclined plane with angle 30 from the horizontal. The block has an initial velocity of 5 (from pushing or something, I forget), and the kinetic friction constant is 0.4. What distance does it travel before it stops?

I have no idea why they put that height of 10m in the figure. I didn't use it at all.

Homework Equations



The Attempt at a Solution


I did my free-body-diagram like so:
14mgwt5.jpg

(I forgot to add to that image, I made positive x to the right, and positive y up)

Then I went to work on Newtons second law to see what I could get.

ƩF_{x} = ma

0 - mgsin(30) - f_{k} = ma

-mgsin(30) - \mu_{k}N = ma

-mgsin(30) - \mu_{k}mgcos(30) = ma

-mgsin(30) - (0.4)mgcos(30) = ma

-gsin(30) - (0.4)gcos(30) = a //Divided out the mass

-(9.8)sin(30) - (0.4)(9.8)cos(30) = a

-8.29 = a

Then, using kinematics:

\upsilon_{f} = \upsilon_{i} + 2a \Delta x

0 = 5 + 2(-8.29) \Delta x

-5 = 2(-8.29) \Delta x

\frac{-5}{2(-8.29)} = \Delta x

0.302 = \Delta x

So my solution was 0.302 meters. I may have made a mistake re-working this directly in LateX, but if that is the right idea for this problem I should have got that one on the test. Was I incorrect? I also made sure to specify that my delta x was across the hyp of the inclined plane.
 
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I get that answer, too.
 
Shouldn't that kinematic formula be:

{\upsilon_{f}}^2 = {\upsilon_{i}}^2 + 2a \Delta x
 
gneill said:
Shouldn't that kinematic formula be:
Yes, it should be. I didn't check OP's working; I calculated independently.

Amazingly, I overlooked squaring the 5, in effect the same mistake, so my value is out by a factor of 5. My answer is now 1.5m up the slope.

Thanks for pointing out QuarkCharmer's mistake, it made me review my working.
 
Oh yeah, you are right haha. I confused it with the one that is:
v_f = V_i + at

But I am sure I used the correct one on the test (I specifically remember writing 0^2 as one line). I just did this one from what I could remember in the physicsforum.com input box.
 
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