- #1

shmoop

## Homework Statement

Hi! I'm having issues with this practice problem. Any advice on what I'm doing wrong would help!!!

A 2.5 kg particle is projected with an initial speed of 3.3m/s along a surface for which the coefficient of friction is 0.7. Find the distance it travels given that the particle moves up a 26 degree incline.

## Homework Equations

W=ΔKE

W=F*d

Ff=μFn

W=ΔKE+ΔPE

KE=1/2mv

^{2}

PE=mgh

## The Attempt at a Solution

mg=2.5kg*-9.8m/s

^{2}=-24.5N

-24.5Nsin26=-10.74 N = mgx

-24.5Ncos26=-22.02N = mgy

ΣFy=0=Fn-mgy

Fn= 22.02 N

Ff = Fnμ

Ff = 22.02N*0.7

Ff=15.14 N

W= KEf-KEf+PEf-PEi

W=.5mv

^{2}-.5mv

^{2}+mgh-mgi

W=0.5*0.3kg*(3.3m/s)

^{2}-0+2.5kg*9.8*h-0

W=F*d

Here I assumed KEf was 0 because the final velocity is 0, and I assumed PEi was 0 because the inital height is 0.

The forces applied that aren't in equilibrium are mgx, and Ff.

W= (-15.14N-10.74N)*d

-26.15N*d=0.5*0.3kg*(3.3m/s)

^{2}-0+2.5kg*9.8*h

So instead of having 2 unknown variables, I made h equal to...

h= d*sin26

So.. I inserted that in the previous equation:

-26.15N*d=0.5*0.3kg*(3.3m/s)

^{2}-0+2.5kg*9.8*d*sin26

Factored out d...

d(-26.15+24.5sin26)=-13.6125

d=-13.6125/(-26.15+24.5sin16)

d=0.88m

And this answer is incorrect. I am unsure where I went wrong... Any thoughts? Thank you!!!

For an image of what my free body diagram looks like, I posted it below!!

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