91Ó°ÊÓ

Kinetic energy for displacement $x=5.0\text{m}$is$K_0=30.0\text{J}$$m=8.0\text{kg}$

The problem is based on the work energy principle. It can be used to solve problems of general mechanics. According to this principle, the work done by the net force applied to a body is equal to the change of the kinetic energy of the body. Also, it deals with the kinematic equation of motion in which the motion of an object can be described at constant acceleration.

Formula:

$v_f^2=v_0^2+a\mathrm{Δ}x$

Where$v_0$ and $v_f$are the initial and final velocities.

Using equation (i), the change in kinetic energy is given by,

$$\begin{gathered} \mathrm{Δ}K=\frac{1}{2}m\left(v_f^2-v_0^2\right) \\ =\frac{1}{2}m\left(2a\mathrm{Δ}x\right) \\ =ma\mathrm{Δ}x \end{gathered}$$

$$\begin{gathered} a=\frac{\mathrm{Δ}K}{m\mathrm{Δ}x} \\ =\frac{-30\text{J}}{8.0\text{kg}\left(5.0\text{m}\right)} \end{gathered}$$

Thus, $a=-0.75{\text{m/s}}^{\text{2}}$

Here, the negative sign indicates that the mass is decelerating. Also, at $x=5.0\text{m}$the kinetic energy becomes zero. Thus, the mass comes to rest momentarily. Using equation (i), we get the initial velocity as,

$$\begin{gathered} v_0^2=v_f^2-2a\mathrm{Δ}x \\ =0-2\left(-0.75{\text{m/s}}^{\text{2}}\right)\left(5.0\text{m}\right) \\ =7.5{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}} \end{gathered}$$

Using the above value, we get the velocity at $x=-3.0\text{m}$is,

$$\begin{gathered} v_f=\sqrt{v_0^2+2a\mathrm{Δ}x} \\ =\sqrt{7.5{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}+2\left(-0.75{\text{m/s}}^{\text{2}}\right)\left(-3.0\text{m}\right)} \end{gathered}$$

Thus, the velocity when object moves back through $x=-3.0\text{m}$is$3.5\text{m/s}$