This group/cooperative learning problem is designed to see if you can combine two aspects of kinematical motion, construct a valid space time plot and then solve for the requested result.


QUESTION: An object, 2 m above the floor, tethered by a chain which is attached to a brake (a constant force spring) is released and falls vertically under the influence of gravity under it reaches the end of its tether and then is deaccelerated by the ``brake'' which can, in the absence of any other effect (gravity included) has an acceleration of 14.7 m/s2. How long should the tether chain be so that the falling object just ``kisses'' the floor?


As the figure below shows there are two regimes to the fall.

1.
From t0 to t1 in which the object falls a distance x and experiences and acceleration of a= -g.
2.
From t1 to t2 in which time object falls a distance y and experiences the acceleration of gravity -g and the braking deacceleration of 14.7 m/s2 or 3g/2.
\includegraphics [height=3.1in]{pd_fig1.eps}

To solve the problem:

1.
You must utilize the vector aspect of acceleration in the second time regime and find that a'= (3g/2 + -g) = g/2.
2.
Next establish a space-time plot (if you can do this in your mind this is fine.)
3.
Choose which expressions are least cumbersome to use and solve.

\includegraphics [height=4.1in]{pd_fig2.eps}

Of the three motion equations:

\begin{displaymath} x = x_0 + v_0 t + \frac{1}{2} at^2 \end{displaymath}

v = v0 t + a t

v2 = v02 + 2 a (x-x0)

only the last doesn't require a knowledge of the temporal aspect of the problem. Using a space-time to establish the motion and facts you should find that you obtain two useful expressions:
Time regime one: v2 = (0)2 + 2 (-g) x
Time regime two: 0 = v2 + 2 (g/2) y
with $x+y = -2m \equiv s$

Substituting gives 0 = 2 (-g) x + 2 (g/2) (s - x) and solving gives x = s/3 = -0.67 meters.