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Part A

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A person pushes a box of mass 15 kg along a floor by applying a perfectly horizontal force F. The box accelerates horizontally at a rate of 2.0 m/s2. Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of F?

System: Box as point particle subject to external influences from the person (applied force) the earth (gravity) and the floor (normal force and friction).

Model: Point Particle Dynamics.

Approach: The free body diagram for this situation is:

With this free body diagram, Newton's 2nd Law can be written:

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\begin

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[ \sum F_

Unknown macro: {x}

= F - F_

Unknown macro: {f}

= ma_

]
[ \sum F_

Unknown macro: {y}

= N - mg = ma_

= 0 ]\end

where we have assumed that the y acceleration is zero because the box is sliding along a horizontal floor, not moving upward or downward. This realization is important, because we know Ff = μN. Thus, because the y acceleration is zero, we can solve Newton's 2nd Law in the y direction to yield:

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\begin

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[ N = mg]\end

so that:

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\begin

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[ F = ma_

Unknown macro: {x}

+F_

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= ma_

+ \mu_

Unknown macro: {k}

N = ma_

Unknown macro: {x}

+ \mu_

mg = \mbox

Unknown macro: {96 N}

] \end

Part B

A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force F. The box moves horizontally at a constant speed of 2.0 m/s in the direction of the person's applied force. What is the magnitude of F?

System and Model: As in Part A.

Approach: Just as above, Newton's 2nd Law can be written:

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\begin

Unknown macro: {large}

[ \sum F_

Unknown macro: {x}

= F - F_

Unknown macro: {f}

= ma_

]
[ \sum F_

Unknown macro: {y}

= N - mg = 0] \end

This time, however, the x acceleration is also zero, since the box maintains a constant speed. This implies:

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\begin

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[ F = ma_

Unknown macro: {x}

+ F_

Unknown macro: {f}

= F_

= \mu_

Unknown macro: {k}

mg = \mbox

Unknown macro: {66 N}

] \end

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