A force sensor is used to measure a force applied to a 9.1 kg box on a hard, horizontal surface. It records that the box starts to move when the force is equal to 26.9 N. This force makes the box start to move with an acceleration of 0.2 m/s2. Find mu sub s/k respectively.

Answer 1

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

Answer 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