You are viewing an old version of this page. View the current version.
Compare with Current
View Page History
Version 1
Next »
Unknown macro: {table}
Unknown macro: {tr}
Unknown macro: {td}
Error formatting macro: live-template: java.lang.NullPointerException
Unknown macro: {td}
Unable to render embedded object: File (pushingbox.png) not found.
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:
Unknown macro: {latex}
\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
Error formatting macro: live-template: java.lang.NullPointerException