Question

For the arrangement of forces in Problem $81,$ a $2.00kg$ particle is released at $x=5.00m$ with an initial velocity of $3.45m/s$ in the negative direction of the x axis. (a) If the particle can reach $x=0m,$ what is its speed there, and if it cannot, what is its turning point? Suppose, instead, the particle is headed in the positive x direction when it is released at $x=5.00m$ at speed $3.45m/s.$ (b) If the particle can reach $x=13.0m,$ what is its speed there, and if it cannot, what is its turning point?

Answer 1

(a) At $x=5.00m$ the potential energy is zero, and the kinetic energy is

$K=\frac{1}{2}m{v}^2=\frac{1}{2}\left(2.00kg\right)(3.45m/s{)}^2=11.9J.$

The total energy, therefore, is great enough to reach the point $x=0$ where $U=11.0J$,with a little “left over” $(11.9J-11.0J=0.9025J).$. This is the kinetic energy at $x=0$,which means the speed there is

$v=\sqrt{2\left(0.9025J\right)/\left(2kg\right)}=0.950m/s.$

It has now come to a stop, therefore, so it has not encountered a turning point.

(b) The total energy $\left(11.9J\right)$ is equal to the potential energy (in the scenario where it is initially moving rightward) at $x=10.9756\approx 11.0m.$ This point may be found by interpolation or simply by using the work-kinetic energy theorem:

$K_f=K_i+W=0\Rightarrow 11.9025+(-4)d=0\Rightarrow d=2.9756\approx 2.98$

(which when added to $x=8.00$ [the point where $F_3$ begins to act] gives the correct result).This provides a turning point for the particle’s motion.