- #1

QuarkCharmer

- 1,051

- 3

## 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.

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:

(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.

[itex]ƩF_{x} = ma[/itex]

[itex]0 - mgsin(30) - f_{k} = ma[/itex]

[itex]-mgsin(30) - \mu_{k}N = ma[/itex]

[itex]-mgsin(30) - \mu_{k}mgcos(30) = ma[/itex]

[itex]-mgsin(30) - (0.4)mgcos(30) = ma[/itex]

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

[itex]-(9.8)sin(30) - (0.4)(9.8)cos(30) = a[/itex]

[itex]-8.29 = a[/itex]

Then, using kinematics:

[itex]\upsilon_{f} = \upsilon_{i} + 2a \Delta x[/itex]

[itex]0 = 5 + 2(-8.29) \Delta x[/itex]

[itex]-5 = 2(-8.29) \Delta x[/itex]

[itex]\frac{-5}{2(-8.29)} = \Delta x[/itex]

[itex]0.302 = \Delta x[/itex]

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.