For the arrangement of forces in Problem 81, a 2.00 kgparticle is released at x = 5.00mwith an initial velocity of 3.45 m/sin the negative direction of the xaxis. (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 xdirection when it is released at x = 5.00 mat speed 3.45 m/s. (b) If the particle can reach x = 13.0 m, what is its speed there, and if it cannot, what is its turning point?

The mass of the particle is 2.00 kg

The initial position of the particle is x = 5.00m

The initial velocity of the particle is u = 3.45 m/s

The potential energy at x = 0 m is U = 11 J (Obtained from Problem 81)

Using the work-energy theorem, we can solve this problem.

${\mathbf{K}}_{\mathbf{i}}\mathbf{+}{\mathbf{W}}_{\mathbf{i}}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{+}{\mathbf{W}}_{\mathbf{f}}$

At x = 5.00 m the potential energy is zero, and the kinetic energy can be expressed as,

$\mathrm{K}.\mathrm{E}=\frac{1}{2}{\mathrm{mu}}^2$

Substituting the given data in the above expression, we get,

role="math" localid="1661405210389" $$\begin{gathered} K.E.=\frac{1}{2}m{u}^2 \\ =\frac{1}{2}(2.00\mathrm{kg}){(3.45\mathrm{m}/\mathrm{s})}^2 \\ =11.9\mathrm{kg}.{\mathrm{m}}^2.{\mathrm{s}}^{-2}\left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2.{\mathrm{s}}^{-2}}\right) \\ =11.9\mathrm{J} \end{gathered}$$

The total energy of the particle is, therefore, great enough to reach the point x = 0m where U = 11.0 J.

Since the particle overcomes this barrier then the leftover energy of the particle will be the kinetic energy over this barrier, such that,

$$\begin{gathered} K.E_{leftover}=11.9\mathrm{J}-11.0\mathrm{J} \\ =0.9\mathrm{J} \end{gathered}$$

This is the kinetic energy at x = 0m, which means the velocity of the particle will be,

$$\begin{gathered} \mathrm{K}.{\mathrm{E}}_{\mathrm{leftover}}=0.9\mathrm{J} \\ \frac{1}{2}{\mathrm{mv}}^2=0.9\mathrm{J} \\ \mathrm{v}=\sqrt{\frac{2\times (0.9\mathrm{J})}{2.00\mathrm{kg}}} \\ \mathrm{v}=0.95\mathrm{m}/\mathrm{s} \end{gathered}$$

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

The total energy (11.9J) is equal to the potential energy (in the scenario where it is

initially moving rightward) at x = 11.0 m.

This point may be found by interpolation or simply by using the work-kinetic energy theorem,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{i}}+\mathrm{W}=0 \\ 11.9\mathrm{J}+(-4\mathrm{N})\mathrm{d}=0 \\ \mathrm{d}=\frac{11.9\mathrm{J}}{4\mathrm{N}} \\ \mathrm{d}=2.98\mathrm{m} \end{gathered}$$

(which when added to x = 8.00m [the point where $F_3$ begins to act] gives the correct result).

Therefore, the particle will not be able to reach up to x = 13.0 m

This provides a turning point for the particle’s motion and turns at x = 11.0 m