Work and frictional force on a Sliding Box

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SUMMARY

The discussion centers on calculating the average frictional force acting on a sliding box and the work done by a person pushing the box after it comes to rest. In Part A, the average frictional force, F_f, is derived as F_f = (mv_0^2)/(2x_1), where m is the mass, v_0 is the initial speed, and x_1 is the distance traveled before coming to rest. In Part B, the work done by the person, W_p, is expressed as W_p = (delta)K - W_f, where (delta)K represents the change in kinetic energy. The challenge lies in determining the work done against friction, W_f, which remains constant from Part A to Part B.

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  • Knowledge of work-energy principles
  • Basic algebra for manipulating equations
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of motion involving friction and work-energy principles.

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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