Posted by Brittney on Monday, June 18, 2012 at 10:19am.
F(fr) =μ•N = μ•m•g
1. F(fr)=F
μ(s)•m•g =F,
μ(s) =F/m•g =26.9/9.1•9.8 = 0.3
2. m•a=F(fr)
m•a = μ(k)•m•g
μ(k)=a/g=0.2/9.8=0.2
First compute mu(static):
When the box just begins to move, the applied force equals mu(s)*m*g
So mu(s)= F/m*g = 26.9/(9.1*9.8)= 0.30
Now when the box gets into motion, mu(k) comes in picture and frictional force f = mu(k)*m*g
So F - mu(k)*m*g = m*a
mu(k) = (F - m*a)/m*g
= (26.9-9.1*0.2)/9.1*9.8
= 0.28
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