Work and frictional force on a Sliding Box

[danni7070]

Homework Statement

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

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.

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.

Homework Equations

My solution in part A is that Ff = (mv_0^2)/(2x_1)

The Attempt at a Solution

But my problem is in Part B

I know that W_p = (delta)K - W_f

Where (delta)K = K_final (because K_initial = 0)

My problem is finding W_f and it is given that F_f is the same in Part A as in Part B...

Please help, as I'm to blind to see the probably obvious solution, and if you don't understand something what I'm writing, please ask because my english is not that good.

Thanks.

 

[danni7070]

And yes I also know that K_final is 0.5(mv_1^2)

 

[m2003]

I would like to first clarify that W_p is the work done by the person pushing the box, and W_f is the work done by the frictional force.

In Part B, we can first calculate the work done by the frictional force, which is the same magnitude as in Part A. This would be W_f = F_f * (x_2 - x_1), where F_f is the average frictional force calculated in Part A and (x_2 - x_1) is the distance over which it acts.

Next, we can calculate the change in kinetic energy of the box, which is equal to the work done by the person pushing it. This would be (delta)K = (1/2) * m * (v_1^2 - v_0^2).

Therefore, the work done by the person pushing the box would be W_p = (1/2) * m * (v_1^2 - v_0^2) - F_f * (x_2 - x_1).

In terms of the given variables, this would be W_p = (1/2) * m * (v_1^2 - v_0^2) - [(mv_0^2)/(2x_1)] * (x_2 - x_1).

I hope this helps clarify the solution for Part B. Let me know if you have any further questions or need any further explanation.