very easy(only part B required)?

Question

A box of mass m is sliding along a horizontal surface.

Part A:(solved)


The box leaves position x = 0 with speed v_0. The box is slowed by a constant frictional force until it comes to rest at position x = x_1.

Find F_f, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.)
Express the frictional force in terms of m, v_0, and x_1.
I figured out answer to be: 0.5*m*v0^2/x1
Part B:
After the box comes to rest at position x_1, a person starts pushing the box, giving it a speed v_1.

When the box reaches position x_2 (where x_2 > x_1), how much work W_p has the person done on the box?
Assume that the box reaches x_2 after the person has accelerated it from rest to speed v_1.
Express the work in terms of m, v_0, x_1, x_2, and v_1.
It has something to do with conservation of energy

Answer 1

The work done in part two is going to be the work required to get the box moving at v_1, plus the frictional force overcome pushing it that distance. You found the force of friction so that's
Ff*distance=
(.5m*v0^2/x_1)(x_2 - x_1)
This would be the amount of work if you just moved the box along the floor at constant speed. But you also speed up the box by doing .5mv_1^2 extra work, or the total work done is:

(.5m*v0^2/x_1)(x_2 - x_1) + .5mv_1^2

Answer 2

Part A Answer is Ff = mv0^(2)/2x_1 (no .5 added)

Answer 3

Part A Answer is Ff = .5mv02/x1

Answer 4

[{(mv_0^2/2) * (x_2-x_1)}/x_1] + mv_1^2/2