Part A
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/s2. What is the magnitude of F?
Solution
System:
Box as point particle.
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 for the x direction can be written:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ \sum F_
Unknown macro: {x}
= F = ma_
= \mbox
Unknown macro: {30 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?
Solution
System, Interactions and Model:
As in Part A.
Approach:
Just as above, Newton's 2nd Law for the x direction can be written:
Unknown macro: {latex}
\begin
Unknown macro: {large}
[ \sum F_
Unknown macro: {x}
= F = ma_
] \end
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:
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\begin
Unknown macro: {large}
[ F = ma_
Unknown macro: {x}
= \mbox
Unknown macro: {(15 kg)(0 m/s}
^
Unknown macro: {2}
) = \mbox
Unknown macro: {0 N}
] \end