# Problem: A box is sliding with a constant speed of 3.80  m/s in the +x-direction on a horizontal, frictionless surface. At x = 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between x = 0 and x = 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400.(a) What is the x-coordinate of the point where the box comes to rest?(b) How much time does it take the box to come to rest after it first encounters the rough patch at x = 0?

###### FREE Expert Solution

We're asked for the x-coordinate (position) where an object acted on by friction comes to rest and the time it takes it takes to come to rest.

This is a Friction type of problem with kinematics. When we're tackling a problem that has forces and kinematics, we'll follow these steps:

1. Draw free body diagrams (FBDs)
2. Set up Newton's 2nd Law equations ( Fma )
3. Set up (choose) kinematics equations
4. Solve for the target variables

Remember that we calculate kinetic friction using

$\overline{){{\mathbit{f}}}_{{\mathbit{k}}}{\mathbf{=}}{{\mathbit{\mu }}}_{{\mathbit{k}}}{\mathbit{N}}}$

The four UAM (kinematics) equations are:

93% (299 ratings) ###### Problem Details

A box is sliding with a constant speed of 3.80  m/s in the +x-direction on a horizontal, frictionless surface. At x = 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between x = 0 and x = 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400.

(a) What is the x-coordinate of the point where the box comes to rest?

(b) How much time does it take the box to come to rest after it first encounters the rough patch at x = 0?