Answer:
[tex]w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)[/tex]
Explanation:
First, the box goes from position x=0 to x=x1, so, the work made by the friction is calculated as:
[tex]-F_f(x_1) = \frac{1}{2}mV^2 -\frac{1}{2}mv_0^2[/tex]
where [tex]F_f[/tex] is the force of friction, [tex]x_1[/tex] the displacement, m the mass, V is the speed in rest and [tex]v_0[/tex] the initial speed.
V is equal to zero, so:
[tex]F_f(x_1)=\frac{1}{2}mv_0^2[/tex]
it means that the force of friction [tex]F_f[/tex] is equal to:
[tex]F_f = \frac{1mv_0^2}{2x_1}[/tex]
On the other hand, when the box goes from position x=x1 to x=x2, the work-energy theorem said that:
[tex]-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2-\frac{1}{2}mV^2[/tex]
Where [tex]w_p[/tex] is the work done by the person.
Then, solving for [tex]w_p[/tex]:
[tex]-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2[/tex]
[tex]w_p = \frac{1}{2}mv_1^2+F_f(x_1-x_2)[/tex]
Finally, replacing [tex]F_f[/tex] by [tex]\frac{1mv_0^2}{2x_1}[/tex], we get:
[tex]w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)[/tex]
