A `5 kg ` box is moving straight across the floor at a constant speed by a force of `30 N` as shown in figure

a. How large a friction force impends the motion of the box ?
b. Find `mu_(k)` between the box and the floor

Explanation

`N = mg - 30 sin30^(@)`
As the speed is constant so the acceleration is zero
a. Hence friction `f_(k) = 30 cos 30^(@) = 15sqrt(3) N`
b.`f_(k) = 30 cos 30^(@) = mu_(k) (mg - 30 sin 30^(@))`
Solving we get `mu_(k) = 3sqrt(3//7)`