This Example is incomplete.
PUSHING A BOX
In this scenario a person applies a constant force to push a box across a plane which is assumed smooth (meaning frictionless). To start the motion with a constant force, imagine the box restrained by a "shear-pin" (a steel post, inserted into the floor). When the person pushes the box with increasing applied force, the shear-pin holds until the force on it becomes 50 Newtons, whereupon it breaks, releasing the box to horizontal motion. We set the event clock to start the instant the pin shears; that time is t = 0+.
a) As a prelimanary calculation, what is the applied force (a vector) when the force on the shear pin has become 50N? ANS: 100N
b) Above, the sum of forces equaled zero and nothing changed. But once the shear pin broke, the sum was is no longer zero. Below is shown how the differential equation that governs the motion for all times greater than 0+ is obtained.
As the event procedes, time, velocity and position change. Since Newton's Third Law of Motion is a first order differential equation, one final condition is needed for each final state. Solve the differential equation to determine the state of the box for each of the second conditions:
After 3 seconds has lapsed, what is the velocity and position?
After traveling 3 feet, what is the second velocity and time?
When the velocity is 3ft/s, what is the position and time?