Solved Problem on Dynamics
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A body of mass 3 kg, initially at rest, is on a frictionless horizontal surface. A horizontal force of constant magnitude equal to 4.5 N acts on the body for 20 s. Calculate:
a) What is the acceleration of the body while the force acts?
b) What is the speed of the body when the force stops acting?
c) What is the distance traveled by the body until the force stops acting?


Problem data:
  • Mass of the body:    m = 3 kg;
  • Initial speed of the body:    v0 = 0;
  • Force applied to the body:    F = 4.5 N;
  • Interval of time in what the force acts on the body:       t = 20 s.
Problem diagram:

We choose a reference frame oriented to the right, with the body initially at rest at the origin (Figure 1).

Figure 1

Drawing a Free Body Diagram, we have the forces acting on the block (Figure 2).

  • Horizontal direction
    • \( \vec F \) : force applied to the body.
  • Vertical direction
    • \( \vec N \) : normal reaction force of the surface on the body;
    • \( \vec W \) : weight of the body.
Figure 2

Solution

a) Applying Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \end{gather} \]
In the vertical direction, there is no motion, normal force and weight cancel out.
In the horizontal direction, the only force is the force applied to the body, which will be the resultant force in this direction, in magnitude
\[ \begin{gather} a=\frac{F}{m}\\[5pt] a=\frac{4.5\;\mathrm{\frac{\cancel{kg}.m}{s^2}}} {3\;\mathrm{\cancel{kg}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {a=1.5\;\mathrm{m/s^2}} \end{gather} \]

b) As the body has a constant acceleration it is in Uniformly Accelerated Rectilinear Motion, applying the equation of velocity as a function of time
\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_0+at} \end{gather} \]
\[ \begin{gather} v=0+\left(1.5\;\mathrm{\frac{m}{s^{\cancel 2}}}\right)\left(20\;\mathrm{\cancel s}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {v=30\;\mathrm{m/s}} \end{gather} \]

c) Applying the equation of displacement as a function of time to the Uniformly Accelerated Rectilinear Motion
\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_0+v_0t+\frac{a}{2}\;t^2} \end{gather} \]
\[ \begin{gather} S=0+0\times\left(20\;\mathrm s\right)+\frac{1}{2}\left(1.5\;\mathrm{\frac{m}{s^2}}\right)\left(20\;\mathrm s\right)^{2}\\[5pt] S=\frac{1}{\cancel 2}\left(1.5\;\mathrm{\frac{m}{\cancel{s^2}}}\right)\left(\cancelto{200}{400}\;\mathrm{\cancel{s^2}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {S=300\;\mathrm m} \end{gather} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .