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Goozombies
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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]+mv=(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...