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Part A

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.

Solution

The free body diagram for this situation is:

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

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

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 = μN. Thus, because the y acceleration is zero, we can solve Newton's 2nd Law in the y direction to yield:

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

so that:

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

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?