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h4. Part A

!Pushing a Box^pushingbox.png|width=40%!

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/s{color:black}^2^{color}.  {excerpt}Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of _F_?{excerpt}

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:

!pushingboxfrfbd1.png!

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

{latex}\begin{large}\[ \sum F_{x} = F - F_{f} = ma_{x} \]
\[ \sum F_{y} = N - mg = ma_{y} = 0 \]\end{large}{latex}

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 _F_~f~ = μ_N_.  Thus, because the _y_ acceleration is zero, we can solve Newton's 2nd Law in the _y_ direction to yield:

{latex}\begin{large}\[ N = mg\]\end{large}{latex}

so that:

{latex}\begin{large}\[ F = ma_{x}+F_{f} = ma_{x} + \mu_{k}N = ma_{x} + \mu_{k}mg = \mbox{96 N} \] \end{large}{latex}

h4. Part B

A person pushes a box of mass 15 kg along a 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.  Assuming the coefficient of kinetic friction between the box and the ground is 0.45, what is the magnitude of _F_?

System and Model:  As in Part A.

Approach:  Just as above, [Newton's 2nd Law|Newton's Second Law] can be written:

{latex}\begin{large}\[ \sum F_{x} = F - F_{f}= ma_{x} \]
\[ \sum F_{y} = N - mg = 0\] \end{large}{latex}

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

{latex}\begin{large}\[ F = ma_{x} + F_{f} = F_{f} = \mu_{k}mg = \mbox{66 N} \] \end{large}{latex}