Solved Problem on Dynamics

A block of mass 5 kg, is launched with an initial speed of 20 m/s in ascending direction over an inclined plane of 45°. The coefficient of friction between the block and the plan is equal to 0.4. 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°;
Coefficient of friction: μ = 0.4;
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 the block (Figure 2).

\( \vec W \): weight of block;
\( \vec N \): normal reaction force, surface reaction on block;
\( {\vec F}_f \): force of friction between the block and the plane.

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 3-A).
Plotting the forces in an xy coordinate system (Figure 3-B).

Figure 2

Solution

Applying Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec F=m\vec a} \end{gather} \]

Direction x:

\[ \begin{gather} -W_{\small P}-F_f=ma \tag{I} \end{gather} \]

The parallel component of the weight is given by

\[ \begin{gather} W_{\small P}=W\sin\theta \tag{II} \end{gather} \]

substituting equation (II) into equation (I)

\[ \begin{gather} -W\sin\theta-F_f=ma \tag{III} \end{gather} \]

Direction y:

There is no motion in this direction, the normal force and the normal weight component cancel our each other

\[ \begin{gather} N=W_{\small N} \tag{IV} \end{gather} \]

The normal weight component is given by

\[ \begin{gather} W_{\small N}=W\cos \theta \tag{V} \end{gather} \]

substituting equation (V) into equation (IV)

\[ \begin{gather} N=W\cos \theta \tag{VI} \end{gather} \]

The weight is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {W=mg} \tag{VII} \end{gather} \]

substituting the equation (VII) into equations (III) and (VI)

\[ \begin{gather} -mg\sin\theta-F_f=ma \tag{VIII} \end{gather} \]

\[ \begin{gather} N=mg\cos \theta \tag{IX} \end{gather} \]

The friction force is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {{\vec F}_f=\mu \vec N} \tag{X} \end{gather} \]

substituting the equation (X) into equation (VIII)

\[ \begin{gather} -mg\sin\theta-\mu N=ma \tag{XI} \end{gather} \]

substituting equation (IX) into equation (XI)

\[ \begin{gather} -\cancel{m}g\sin\theta -\mu \cancel{m}g\cos \theta=\cancel{m}a\\[5pt] a=-g(\sin\theta +\mu \cos \theta ) \end{gather} \]

the negative sign of the acceleration indicates that its motion is in the opposite direction to the orientation of the trajectory, and the speed of the block is decreasing.

\[ \begin{gather} a=-\left(9.8\;\mathrm{m/s^2}\right)\times(\operatorname{sen}45°-0.4\cos 45°) \end{gather} \]

From Trigonometry we have \( \sin45°=\dfrac{\sqrt{2\;}}{2} \) e \( \cos 45°=\dfrac{\sqrt{2\;}}{2} \)

\[ \begin{gather} a=-\left(9.8\;\mathrm{m/s^2}\right)\times\left(\frac{\sqrt{2\;}}{2}+0.4\times\frac{\sqrt{2\;}}{2}\right)\\[5pt] a=-9.7\;\mathrm{m/s^2} \end{gather} \]

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

\[ \begin{gather} \Delta S=\frac{v^2-v_0^2}{2a}\\[5pt] \Delta S=\frac{\left(0\;\mathrm{m/s}\right)^2-\left(20\;\mathrm{m/s}\right)^2}{2\times\left(-9.7\;\mathrm{m/s^2}\right)}\\[5pt] \Delta S=\frac{-400\;\mathrm{m^{\cancel 2}/\cancel{s^{2}}}}{-19.4\;\mathrm{\cancel m}/\cancel{s^2}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\Delta S=20.6\;\mathrm m} \end{gather} \]

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