Work and frictional force on a Sliding Box

AI Thread Summary
A box of mass m slides along a horizontal surface, initially moving with speed v_0 and is slowed by a constant frictional force until it stops at position x_1. The average frictional force, F_f, is determined to be (mv_0^2)/(2x_1). After stopping, a person pushes the box to accelerate it to speed v_1, and the goal is to calculate the work done, W_p, as the box moves to position x_2. The work done is expressed as W_p = (delta)K - W_f, where (delta)K is the change in kinetic energy, and W_f is the work done by friction. The user seeks clarification on calculating W_f, noting that the frictional force remains the same as in Part A.
danni7070
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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.
 
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And yes I also know that K_final is 0.5(mv_1^2)
 
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