Question

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. When the box reaches position x2 (where x2>x1), how much work Wp has the person done on the box? Assume that the box reaches x2 after the person has accelerated it from rest to speed v1. Express the work in terms of m, v0, x1, x2, and v1.

Answer 1

Answer:

$w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)$

Explanation:

First, the box goes from position x=0 to x=x1, so, the work made by the friction is calculated as:

$-F_f(x_1) = \frac{1}{2}mV^2 -\frac{1}{2}mv_0^2$

where $F_f$ is the force of friction, $x_1$ the displacement, m the mass, V is the speed in rest and $v_0$ the initial speed.

V is equal to zero, so:

$F_f(x_1)=\frac{1}{2}mv_0^2$

it means that the force of friction $F_f$ is equal to:

$F_f = \frac{1mv_0^2}{2x_1}$

On the other hand, when the box goes from position x=x1 to x=x2, the work-energy theorem said that:

$-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2-\frac{1}{2}mV^2$

Where $w_p$ is the work done by the person.

Then, solving for $w_p$:

$-F_f(x_1-x_2)+w_p = \frac{1}{2}mv_1^2$

$w_p = \frac{1}{2}mv_1^2+F_f(x_1-x_2)$

Finally, replacing $F_f$ by $\frac{1mv_0^2}{2x_1}$, we get:

$w_p = \frac{1}{2}mv_1^2+\frac{1mv_0^2}{2x_1}(x_1-x_2)$