Did I solve this inclined plane problem correctly?

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Homework Help Overview

The discussion revolves around a physics problem involving a block on an inclined plane with a specified angle and initial velocity, where the effects of kinetic friction are also considered. The original poster seeks validation of their solution regarding the distance the block travels before stopping.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster outlines their approach using free-body diagrams and Newton's second law, leading to a kinematic equation to find the distance traveled. Some participants question the correctness of the kinematic formula used, while others independently calculate and find discrepancies in their results.

Discussion Status

Participants are actively engaging with the original poster's calculations, identifying potential errors in the application of kinematic equations. There is a recognition of mistakes made by multiple individuals, prompting further review of their work without reaching a definitive consensus on the correctness of the original poster's solution.

Contextual Notes

Participants note the presence of an additional height measurement in the problem statement, which the original poster did not utilize in their calculations. This aspect raises questions about its relevance to the solution.

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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