Toy Track

Apply Conservation of Energy to each point, then consider the height in each case to solve for the velocity.

The equation for Conservation of Energy can be written as:

\begin{equation*}
E_i=E_X,
\end{equation*}

where \(X\) represents any point. Conservation of Energy can then be written as:

\begin{equation*}
\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_X^2+mgh_X,
\end{equation*}

where \(v_i=0\) because the car is released from rest. Solving for \(v_X\), we get:

\begin{equation*}
v_X=\sqrt{2g(h_i-h_X)}.
\end{equation*}

Since \(h_i=0.8\,\text{m}\), for \(A\) with \(h_A=0\,\text{m}\), we get:

\begin{equation*}
v_A \approx 3.96 \, \text{m/s}.
\end{equation*}

For \(B\) with \(h_B=0.5\,\text{m}\), we get:

\begin{equation*}
v_B \approx 2.42 \, \text{m/s}.
\end{equation*}

For \(C\) with \(h_C=0.2\,\text{m}\), we get:

\begin{equation*}
v_C \approx 3.43 \, \text{m/s}.
\end{equation*}

For \(D\) with \(h_D=0.6\,\text{m}\), we get:

\begin{equation*}
v_D \approx 1.98 \, \text{m/s}.
\end{equation*}

For \(E\) with \(h_E=0.2\,\text{m}\), we get:

\begin{equation*}
v_E \approx 3.43 \, \text{m/s}.
\end{equation*}

The problem is inquiring about the speed of the car on each of the points shown in the figure. To find that speed, we will use that the friction is negligible, and so the mechanical energy of the car is conserved. That is, the mechanical energy \(E\) at all points in question must be the same. For this case, the mechanical energy will be the sum of the kinetic energy \(K\) and the gravitational potential energy \(U_g\). Thus, for any point we can write

\begin{equation}
\label{mecanica}
E=K+U.
\end{equation}

The kinetic energy can be expressed in terms of the mass \(m\) and speed \(v\) of the toy-car as

\begin{equation}
\label{kinetic}
K=\frac{1}{2}mv^2.
\end{equation}

Figure 1: Heights of each point of the movement of the car.

The gravitational potential energy is given in terms of the mass \(m\), the gravitational acceleration on earth \(g\) and the height \(h\) measured with respect to some reference point. Our reference point will be such that the height of point A is zero, as shown in figure 1.

The coordinate system is placed on the ground at point A.

The explicit expression for the gravitational potential energy is

\begin{equation}
\label{grav}
U_g=mgh.
\end{equation}

Using the kinetic and gravitational potential energy given by equations \eqref{kinetic} and \eqref{grav} into equation \eqref{mecanica}, we obtain

\begin{equation}
\label{emec}
E=\frac{1}{2}mv^2+mgh.
\end{equation}

From now on, we will use a subscript with the name of the point to indicate that this variable is measured at such point. For example, the height at A is \(h_A=0\,\text{m}\). We will name the initial point with the letter \(i\).

Now, in order to calculate the speed for each point we must use the fact that the mechanical energy is conserved, meaning that the initial mechanical energy \(E_i\) is the same as the energy for any other point:

\begin{equation}
E_i=E_X,
\end{equation}

where \(X\) is used to express an arbitrary point, for example, point \(D\). Using the explicit expression for the mechanical energy given by equation \eqref{emec} for both sides of the equation \(E_i=E_X\), we get

\begin{equation}
\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_X^2+mgh_X.
\end{equation}

Now we cancel out the mass \(m\) since it is present in all the terms, and we also use that \(v_i=0\) because the car is dropped with no initial speed. Doing this, we get

\begin{equation}
gh_i=\frac{1}{2}v_X^2+gh_X.
\end{equation}

Solving for \(v_X\), we obtain

\begin{equation}
\frac{1}{2}v_X^2=g(h_i-h_X).
\end{equation}

After multiplying both sides by two, this becomes

\begin{equation}
v_X^2=2g(h_i-h_X),
\end{equation}

and after taking the square-root on both sides we finally get

\begin{equation}
v_X=\sqrt{2g(h_i-h_X)},
\end{equation}

which gives us the speed at any point \(X\) in terms of the initial height and the height of the corresponding point.  For point A we have

\begin{equation}
v_A=\sqrt{2g(h_i-h_A)}.
\end{equation}

Use the numerical values, noticing from the last figure the heights, in SI units, of \(h_i=0.8\,\text{m}\) and \(h_A=0\,\text{m}\). We get

\begin{equation}
v_A=\sqrt{2(9.8\,\text{m/s}^2)(0.8\,\text{m}-0\,\text{m})},
\end{equation}

\begin{equation}
v_A\approx 3.96\,\text{m/s}.
\end{equation}

For point B we have

\begin{equation}
v_B=\sqrt{2g(h_i-h_B)}.
\end{equation}

Again, from the figure we see that, in SI units,  \(h_B=0.5\,\text{m}\) and so we get

\begin{equation}
v_B=\sqrt{2(9.8\,\text{m/s}^2)(0.8\,\text{m}-0.5\,\text{m})},
\end{equation}

\begin{equation}
v_B\approx 2.42\,\text{m/s}.
\end{equation}

For point C we write

\begin{equation}
v_C=\sqrt{2g(h_i-h_C)}.
\end{equation}

From the last figure we see that \(h_C=0.2\,\text{m}\), and so we obtain

\begin{equation}
v_C=\sqrt{2(9.8\,\text{m/s}^2)(0.8\,\text{m}-0.2\,\text{m})},
\end{equation}

\begin{equation}
v_C\approx 3.43\,\text{m/s}.
\end{equation}

For point D we can write

\begin{equation}
v_D=\sqrt{2g(h_i-h_D)}.
\end{equation}

Once more, using the last figure, we find that \(h_D=0.6\,\text{m}\) and so the result is

\begin{equation}
v_D=\sqrt{2(9.8\,\text{m/s}^2)(0.8\,\text{m}-0.6\,\text{m})},
\end{equation}

\begin{equation}
v_D\approx 1.98\,\text{m/s}.
\end{equation}

Finally, for point E, we have

\begin{equation}
v_E=\sqrt{2g(h_i-h_E)}.
\end{equation}

The corresponding height is \(h_E=0.2\,\text{m}\), and so the speed is

\begin{equation}
v_E=\sqrt{2(9.8\,\text{m/s}^2)(0.8\,\text{m}-0.2\,\text{m})},
\end{equation}

\begin{equation}
v_E\approx 3.43\,\text{m/s}.
\end{equation}

As we can clearly see from these results, as the height increases and gets nearer \(h_i\), the speed decreases. This is expected, since the higher the car goes, the higher the gravitational potential energy and so the lesser the kinetic energy (so that the total mechanical energy is conserved).

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