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A baggage handler at an airport applies a constant horizontal force with magnitude F1 to push a box, of mass m, across a rough horizontal surface with a very small constant acceleration a.
The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a.
How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?

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MechEngineer

Answer:

[tex]F_2 = 2F_1 - F[/tex] where F is the friction force that is the same for both boxes

Explanation:

So the same boxes at the same mass m would create a same friction force, let this force be F. The net force in the first and 2nd cases would be

- 1st case: F1 - F

- 2nd case F2 - F

According to Newton 2nd law, this net would make acceleration

- 1st case [tex]a_1 = (F_1 - F)/m[/tex]

- 2nd case[tex]a_2 = (F_2 - F)/m[/tex]

Since [tex]a_2 = 2a_1[/tex]

[tex](F_2 - F)/m = 2(F_1 - F)/m[/tex]

[tex]F_2 - F = 2F_1 - 2F[/tex]

[tex]F_2 = 2F_1 - F[/tex]

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