(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

You and your friend Peter are putting new shingles on a roof pitched at 21^\circ . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.4 m away, asks you for the box of nails. Rather than carry the 2.4 kg box of nails down to Peter, you decide to give the box a push and have it slide down to him.

If the coefficient of kinetic friction between the box and the roof is 0.52, with what speed should you push the box to have it gently come to rest right at the edge of the roof?

2. Relevant equations

f=ma

f(kinetic friction)= NMk

3. The attempt at a solution

Ok..here is what I did:

I drew a FBD of the mass, and got the following:

mg pointing straight down in negative g direction, N diagnaly up to the left, andMk diagnaly up to the right

Then I got there components as follows ( I think I may have used the wrong angles):

for the normal force: N(cos45i-sin45j)

Fgrav: -mgj

F(friction):MkN(-cos 69+sin69)

( I used 69 for the angle because I made a right triangle from the roof at angle 21, and then subtracted 21 from 90 to get the angle to get the components of friction.)

then for forces in the i direction I have: Ncos45-MkN69=ma

a= Ncos45-Mkn69/m

for forces in the j direction I got:

MkNsin69-Nsin45-mg=0

MkN-Nsin45=mg

N(Mksin69-sin45)=mg

N=mg/Mksin69-sin45

N=(2.4)(9.8)/.52sin69-sin45

N= -106

I plugged N into the equation for a and got:

-106(cos45)-.52(-106)cos69/2.4

a=-14.73

Then to find the initial velocity I used the kinematics equation: V^2= V(i)^2+2a(delta)x

0=V(i)^2+2(-14.73)(5.4)

V(i)+ (159.084)^1/2

V(i)= 12.61 m/s

Hopefully this isn't to hard to follow...and someone can tell me where I went wrong!

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# Homework Help: Find the intial velocity given the coefficient of kinetic friction

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