Solved Problem on Dynamics

A body with a mass of 2 kg and an initial velocity of 10 m/s in the positive direction is under the action of a force that varies with position, given by

\[ \begin{gather} F_x=-8x \qquad\qquad\text{units (SI)} \end{gather} \]

What will be the distance traveled by this body until its velocity becomes zero?

Problem data:

Mass of the body: m = 2 kg;
Initial velocity of the body: v₀ = 10.

Problem diagram:

We choose a reference frame with the x-axis pointed to the right and the y-axis upward (Figure 1).

Figure 1

The given force is negative, it is a resistive force that acts in the opposite direction of motion. The force reduces the body's velocity until its final velocity is zero.

Solution:

Applying Newton's Second Law in the x-direction, the given force in the problem is the only force in this direction.

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathbf F=m\frac{d\mathbf v}{dt}} \end{gather} \]

\[ \begin{gather} F_x=m\frac{dv_x}{dt}\\[5pt] -8x=2\frac{dv_x}{dt} \end{gather} \]

multiplying and dividing the right side of the equation by dx

\[ \begin{gather} -4x=\frac{dv_x}{dt}\frac{dx}{dx} \end{gather} \]

reversing the order of the integration terms in the denominator

\[ \begin{gather} -4x=\frac{dv_x}{dx}\frac{dx}{dt} \end{gather} \]

applying the definition of velocity

\[ \begin{gather} \bbox[#99CCFF,10px] {v=\frac{dx}{dt}} \end{gather} \]

integrating in dx on both sides of the equality

\[ \begin{gather} \int -4x\;dx=\int v_x\frac{dv_x}{dx}\;dx\\[5pt] \int-4x\;dx=\int v_x\;dv_x \end{gather} \]

the constant factor (−4) comes out of the integral on the left side. The limits of integration are from 0, the initial position, and x at any instant in dx, and from 10 m/s, the initial velocity, to 0, the final velocity in dv_x.

\[ \begin{gather} -4\int_0^x x'\;dx'=\int_{10}^0 v'_x\;dv'_x \end{gather} \]

Integral of \( \displaystyle \int_0^x x'\;dx' \)

\[ \begin{gather} \int_0^x x'\;dx'=\left.\frac{x^{1+1}}{1+1}\right|_0^x =\frac{x^2}{2}-\frac{0^2}{2}=\frac{x^2}{2} \end{gather} \]

Integral of \( \displaystyle \int_{10}^0v_x\;dv_x \)

\[ \begin{gather} \int_{10}^0 v_x\;dv_x=\left.\frac{v_x^{1+1}}{1+1}\right|_{10}^0=\frac{0^2}{2}-\frac{10^2}{2}=\frac{100}{2}=50 \end{gather} \]

\[ \begin{gather} -\cancelto{2}{4}\times\frac{x^2}{\cancel 2}=-50\\[5pt] x^2=\frac{-50}{-2}\\[5pt] x=\sqrt{25\;} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {x=5\;\mathrm m} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .