A box of mass M is pushed a distance Δ x across a level floor by a constant applied force F . The coefficient of kinetic friction between the box and the floor is μ . Assuming the box starts from rest, express the final velocity v f of the box in terms of M , Δ x , F , μ , and g .

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aristocles aristocles · Answer 1 · 2020-04-08T20:26:37+02:00

Answer:

Final speed of the box after moving the given distance is

[tex]v = \sqrt{\frac{2(F - \mu Mg)\Delta x}{M}}[/tex]

Explanation:

By work energy theorem we know that work done by all the forces is equal to the change in kinetic energy of the system

Here we know that work is done on the system by external force and friction force

So we will have

[tex]F\Delta x - \mu M g \Delta x = \frac{1}{2}Mv^2 - 0[/tex]

[tex](F - \mu Mg)\Delta x = \frac{1}{2}M v^2[/tex]

so we have

[tex]v = \sqrt{\frac{2(F - \mu Mg)\Delta x}{M}}[/tex]

andromache andromache · Answer 2 · 2022-04-05T17:48:04+02:00

The Final speed of the box after moving the given distance is v = √2(F - μMg) Δx/ M.

Here the work done should be equivalent to the change in the kinetic energy

So

FΔx - μMgΔx = 1/2Mv^2 -

(F - μMg)Δx = 1/2Mv^2

v = √2(F - μMg) Δx/ M.

Hence, The Final speed of the box after moving the given distance is v = √2(F - μMg) Δx/ M.

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