Q16P - Fundamentals Of Physics

Question

An 8.0 kg object is moving in the positive direction of an x axis. When it passes through $x = 0$, a constant force directed along the axis begins to act on it.

Figure 7-29 gives its kinetic energy K versus position x as it moves$x=0$ to $x=5.0$;$K_0= 30.0J$ . The force continues to act. What is v when the object moves back through $x = -3.0 m$?

Step-by-Step Solution

Answer

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

Step 1: Given information:

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

Step 2: Understanding the given information

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:

$v_f^2=v_0^2+a\Delta x$

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

Step 3: To calculate velocity when object moves back through x = - 3 . 0   m .

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

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

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

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,

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

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

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

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