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## Homework Statement

A UPS delivery man pushes a 5kg box (A) along a smooth surface where the coefficient of kinetic friction, [µ][/k], is equal to 0.20. The box slides for 1.5m before colliding with a 3kg box (B). The boxes then move together as one unit on a rough surface with [µ][/k]= 0.9 for 0.8m. Determine the initial velocity of the larger box. This problem was on the test that I took today and I'm not sure if I got it correct, or not.

## Homework Equations

∑F=ma

[F][/fric]=[µ][/k][F][/N]

v[f]^2=v[0]^2+2aΔx

p=p'

g=9.8m/s[2]

## The Attempt at a Solution

∑F[y]=m[A]a[y]

F[N]-F[G]=0

F[N]=F[G]=m[A]g=(5)(9.8)=49N

F[fric]=(.2)(49)=9.8N

∑F[x]=m[A]a[x]

F[fric]=(5)a[x]

a[x]=1.96m/s[2]

v[f]^2=v[0]^2+2aΔx

0=v[0]^2+2(1.96)(-1.5)

v[0][smooth skid]=2.42m/s =v

∑F[y]=m[A+B]a[y]

F[N]-F[G]=0

F[N]=F[G]=m[A+B]g=(5+3)(9.8)=78.4N

F[fric]=(.9)(78.4)=70.56N

∑F[x]=m[A+B]a[x]

F[fric]=(5+3)a[x]

a[x]=8.82m/s[2]

v[f]^2=v[0]^2+2aΔx

0=v[0]^2+2(8.82)(-.8)

v[0][A+B][skid]=3.75m/s=v'

p=p'

m[A]v[A]+m

∑F[y]=m[A+B]a[y]

F[N]-F[G]=0

F[N]=F[G]=m[A+B]g=(5+3)(9.8)=78.4N

F[fric]=(.9)(78.4)=70.56N

∑F[x]=m[A+B]a[x]

F[fric]=(5+3)a[x]

a[x]=8.82m/s[2]

v[f]^2=v[0]^2+2aΔx

0=v[0]^2+2(8.82)(-.8)

v[0][A+B][skid]=3.75m/s=v'

p=p'

m[A]v[A]+m

**v****=(m[A]+m****)v'**

(5)v[A]+(3)(2.42)=(5+3)(3.75)

v[A][beg skid on rough]=4.56m/s

v[f]^2=v[0]^2+2aΔx

(4.56)^2-2(1.96)(-1.5)=v[0][A]^2

v[0][A]=3.86m/s (or something close to that.)

I'm pretty sure I got this one wrong. I wonder how much partial credit I'm going to get...(5)v[A]+(3)(2.42)=(5+3)(3.75)

v[A][beg skid on rough]=4.56m/s

v[f]^2=v[0]^2+2aΔx

(4.56)^2-2(1.96)(-1.5)=v[0][A]^2

v[0][A]=3.86m/s (or something close to that.)

I'm pretty sure I got this one wrong. I wonder how much partial credit I'm going to get...