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h4. Part A

!pushingbox.png|width=40%!

A person pushes a box of mass 15 kg along a smooth floor by applying a perfectly horizontal force _F_.  The box accelerates horizontally at a rate of 2.0 m/s{color:black}^2^{color}.  What is the magnitude of _F_?

h4. Solution

*System:*  Box as [point particle] subject to external .

*Interactions:* External influences from the person (applied force) the earth (gravity) and the floor (normal force).

*Model:* [Point Particle Dynamics].

*Approach:*  The word *smooth* in the problem statement is a keyword, telling us that the floor exerts no horizontal force on the box.  Thus, [Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:

{latex}\begin{large}\[ \sum F_{x} = F = ma_{x} = \mbox{30 N}\] \end{large}{latex}

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h4. 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_?

h4. Solution

*System, Interactions and Model:*  As in Part A.

*Approach:*  Just as above, [Newton's 2nd Law|Newton's Second Law] for the _x_ direction can be written:

{latex}\begin{large}\[ \sum F_{x} = F = ma_{x}\] \end{large}{latex}

This time, however, the acceleration requires some thought.  The speed of the box and its direction of motion are constant.  Thus, by definition, the acceleration is zero.  This implies:

{latex}\begin{large}\[ F = ma_{x} = \mbox{(15 kg)(0 m/s}^{2}) = \mbox{0 N} \] \end{large}{latex}

{info}This result is probably not consistent with your everyday experience.  The reason for this is that it is very difficult to find a box and floor combination with zero friction.  Instead, consider the effort that would be required to keep an air-hockey puck moving at constant speed on the air-table (friction is very small) or to keep a soccer ball rolling at constant speed on a smooth, level floor (friction is unimportant since the ball is rolling).{info}