Question

A box of mass m is sliding along a horizontal surface. part a the box leaves position x=0 with speed v0. the box is slowed by a constant frictional force until it comes to rest at position x=x1. find ff, the magnitude of the average frictional force that acts on the box.

Answer 1

For the work-energy theorem, the work done by the frictional force is equal to the variation of kinetic energy of the box. The initial kinetic energy of the box is
$K_i = \frac{1}{2}mv_0^2$
while the final kinetic energy is zero, because its final speed is zero: 
$K_f =0$
So, the work done by the friction is
$W=K_f-K_i = - \frac{1}{2}mv_0^2$
The negative sign is due to the fact the friction is acting against the direction of the motion. We can neglect it since we only want to find the magnitude of the average frictional force.
We know that the work done by a force is equal to the product between the force and the distance:
$W=Fd$ (1)
In our problem, the distance of the motion is $d=x_1 - 0 = x_1$
So, putting the expression for the work inside (1), and using d=x1 we find
$\frac{1}{2} mv_0^2 = F x_1$
And so, the average frictional force is
$F= \frac{mv_0^2}{2x_1}$