Solved Problem on Dynamics
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A block of mass 5 kg is launched with an initial speed of 20 m/s rising over an inclined plane of 45°. There is no friction between the block and the inclined plane. Determine the distance that the block will come to rest.


Problem data:
  • Mass of the block:    m = 5 kg;
  • Initial speed:    v0 = 20 m/s;
  • Angle of the inclined plane:    θ = 45°;
  • Acceleration due to gravity:    g = 9.8 m/s2;
Problem diagram:

We choose a reference frame pointed out in the upward direction of the inclined plane and with the x-axis parallel to the plane (Figure 1).

Figure 1

Drawing a free-body diagram, we have the forces that act on it (Figure 2-A).

  • \( \vec W \): weight of body;
  • \( \vec N \): normal reaction force of the surface on the body.
The weight can be projected in two directions, a component parallel to the x-axis, \( {\vec W}_{\small P} \), and the other normal or perpendicular, \( {\vec W}_{\small N} \) (Figure 2-A).
Plotting the forces in an xy coordinate system (Figure 2-B)
Figure 2

Solution

Applying Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \end{gather} \]
  • Direction x:
\[ \begin{gather} -W_{\small P}=ma \tag{I} \end{gather} \]
the parallel component of the weight is given by
\[ \begin{gather} W_{\small P}=W\sin\theta \tag{II} \end{gather} \]
the weight is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {W=mg} \tag{III} \end{gather} \]
substituting the equation (III) into equation (II)
\[ \begin{gather} W_{\small P}=mg\sin\theta \tag{IV} \end{gather} \]
substituting the equation (IV) into equation (I)
\[ \begin{gather} -\cancel{m}g\sin\theta=\cancel{m}a\\[5pt] a=-g\sin 45° \end{gather} \]

From trigonometry, we have   \( \sin45°=\dfrac{\sqrt{2}}{2} \)

\[ \begin{gather} a=-\left(9.8\;\mathrm{m/s^2}\right)\times\frac{\sqrt{2\;}}{2}\\[5pt] a\approx -6.9\;\mathrm{m/s^2} \end{gather} \]
the negative sign of the acceleration indicates that its motion is counter to the orientation of the trajectory, and the speed of the block is decreasing.
Using the equation of velocity as a function of displacement
\[ \begin{gather} \bbox[#99CCFF,10px] {v^2=v_0^2+2a\Delta S} \end{gather} \]
The speed of the block decreases until it is equal to zero, v = 0, and substituting the initial speed given in the problem, and the calculated acceleration above
\[ \begin{gather} \Delta S=\frac{v^2-v_0^2}{2a}\\[5pt] \Delta S=\frac{\left(0\;\mathrm{m/s}\right)^2-\left(20\;\mathrm{m/s}\right)^2}{2\times\left(-6.9\;\mathrm{m/s^2}\right)}\\[5pt] \Delta S=\frac{-400\;\mathrm{m^{\cancel 2}/\cancel{s^2}}}{-13.8\;\mathrm{\cancel m/\cancel{s^2}}} \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta S\approx 28.0\;\mathrm m} \end{gather} \]
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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 .