Question

The Total Mechanical Energy of the free falling object at \( \dagger=0 \) s is \[ \begin{aligned} \boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}} & =\boldsymbol{P E}+\boldsymbol{K E} \\ & =432.18+0 \\ & =\mathbf{4 3 2 . 1 8} \mathbf{J} \end{aligned} \] At \( t=1 \mathrm{~s} \), the Potential Energy is, \[ \begin{aligned} \boldsymbol{P E} & =\boldsymbol{m g h} \\ P E & =(1 \mathrm{~kg})(9.8 \mathrm{~m} / \mathrm{s} 2)(44.1 \mathrm{~m}-4.9 \mathrm{~m}) \\ P E & =(9.8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} 2)(39.2 \mathrm{~m}) \\ \boldsymbol{P E} & =\mathbf{3 8 4 . 1 6 \mathbf { J }} \end{aligned} \] The Kinetic Energy at \( t=1 \mathrm{~s} \) is, \[ \begin{aligned} \boldsymbol{K} \boldsymbol{E} & =\frac{\mathbf{1}}{2} \boldsymbol{m} \boldsymbol{v}^{\mathbf{2}} \\ K E & =\frac{1}{2}(1 \mathrm{~kg})(9.8 \mathrm{~m} / \mathrm{s})^{2} \\ \boldsymbol{K} \boldsymbol{E} & =\mathbf{4 8 . 0 2} \boldsymbol{J} \end{aligned} \] The Total Mechanical Energy is: \[ \begin{array}{l} \boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}}=\boldsymbol{P} \boldsymbol{E}+\boldsymbol{K} \boldsymbol{E} \\ M E_{T}=384.16 \mathrm{~J}+48.02 \mathrm{~J} \\ \boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}}=\mathbf{4 3 2 . 1 8} \boldsymbol{J} \end{array} \] Summarizing the answers in Table 1, you can see clearly the equivalence of the Total Mechanical Energy in every second. Following the steps in getting the Kinetic Energy and Potential Energy for \( t=0 \mathrm{~s} \) and \( t=1 \mathrm{~s} \), complete the table when time \( (t)=2 \mathrm{~s} \), and time \( (t)=3 \mathrm{~s} \). Table 1. Summary of the Mechanical Energy of a Free Falling Body \begin{tabular}{|c|c|c|c|c|c|} \hline \( \begin{array}{c}\text { Time, t } \\ \mathbf{( s )}\end{array} \) & \( \begin{array}{c}\text { Height, } \\ \mathbf{h} \\ \mathbf{( m )}\end{array} \) & \( \begin{array}{c}\text { Velocity, } \\ \mathbf{v} \\ (\mathbf{m} / \mathbf{s})\end{array} \) & \( \begin{array}{c}\text { Potential } \\ \text { Energy, PE } \\ (\mathbf{J})\end{array} \) & \( \begin{array}{c}\text { Kinetic } \\ \text { Energy, KE } \\ (\mathbf{J})\end{array} \) & \( \begin{array}{c}\text { Total Mechanical } \\ \text { Energy, } \\ \mathbf{M E}_{\mathrm{T}}=\mathbf{P E + K E} \\ \mathbf{( J )}\end{array} \) \\ \hline 0 & 44.1 & 0 & 432.18 & 0 & 432.18 \\ \hline 1 & 39.2 & 9.8 & 384.16 & 48.02 & 432.18 \\ \hline 2 & & & & & \\ \hline 3 & & & & & \\ \hline \end{tabular} 10

          The Total Mechanical Energy of the free falling object at \( \dagger=0 \) s is
\[
\begin{aligned}
\boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}} & =\boldsymbol{P E}+\boldsymbol{K E} \\
& =432.18+0 \\
& =\mathbf{4 3 2 . 1 8} \mathbf{J}
\end{aligned}
\]
At \( t=1 \mathrm{~s} \), the Potential Energy is,
\[
\begin{aligned}
\boldsymbol{P E} & =\boldsymbol{m g h} \\
P E & =(1 \mathrm{~kg})(9.8 \mathrm{~m} / \mathrm{s} 2)(44.1 \mathrm{~m}-4.9 \mathrm{~m}) \\
P E & =(9.8 \mathrm{~kg} \mathrm{~m} / \mathrm{s} 2)(39.2 \mathrm{~m}) \\
\boldsymbol{P E} & =\mathbf{3 8 4 . 1 6 \mathbf { J }}
\end{aligned}
\]
The Kinetic Energy at \( t=1 \mathrm{~s} \) is,
\[
\begin{aligned}
\boldsymbol{K} \boldsymbol{E} & =\frac{\mathbf{1}}{2} \boldsymbol{m} \boldsymbol{v}^{\mathbf{2}} \\
K E & =\frac{1}{2}(1 \mathrm{~kg})(9.8 \mathrm{~m} / \mathrm{s})^{2} \\
\boldsymbol{K} \boldsymbol{E} & =\mathbf{4 8 . 0 2} \boldsymbol{J}
\end{aligned}
\]
The Total Mechanical Energy is:
\[
\begin{array}{l}
\boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}}=\boldsymbol{P} \boldsymbol{E}+\boldsymbol{K} \boldsymbol{E} \\
M E_{T}=384.16 \mathrm{~J}+48.02 \mathrm{~J} \\
\boldsymbol{M} \boldsymbol{E}_{\boldsymbol{T}}=\mathbf{4 3 2 . 1 8} \boldsymbol{J}
\end{array}
\]
Summarizing the answers in Table 1, you can see clearly the equivalence of the Total Mechanical Energy in every second.

Following the steps in getting the Kinetic Energy and Potential Energy for \( t=0 \mathrm{~s} \) and \( t=1 \mathrm{~s} \), complete the table when time \( (t)=2 \mathrm{~s} \), and time \( (t)=3 \mathrm{~s} \).
Table 1. Summary of the Mechanical Energy of a Free Falling Body
\begin{tabular}{|c|c|c|c|c|c|}
\hline \( \begin{array}{c}\text { Time, t } \\
\mathbf{( s )}\end{array} \) & \( \begin{array}{c}\text { Height, } \\
\mathbf{h} \\
\mathbf{( m )}\end{array} \) & \( \begin{array}{c}\text { Velocity, } \\
\mathbf{v} \\
(\mathbf{m} / \mathbf{s})\end{array} \) & \( \begin{array}{c}\text { Potential } \\
\text { Energy, PE } \\
(\mathbf{J})\end{array} \) & \( \begin{array}{c}\text { Kinetic } \\
\text { Energy, KE } \\
(\mathbf{J})\end{array} \) & \( \begin{array}{c}\text { Total Mechanical } \\
\text { Energy, } \\
\mathbf{M E}_{\mathrm{T}}=\mathbf{P E + K E} \\
\mathbf{( J )}\end{array} \) \\
\hline 0 & 44.1 & 0 & 432.18 & 0 & 432.18 \\
\hline 1 & 39.2 & 9.8 & 384.16 & 48.02 & 432.18 \\
\hline 2 & & & & & \\
\hline 3 & & & & & \\
\hline
\end{tabular}
10

The Total Mechanical Energy of the free falling object at †=0 s is

MET =P E+K E
=432.18+0
=4 3 2 . 1 8𝐉

At t=1 s, the Potential Energy is,

P E =m g h

P E =(1 kg)(9.8 m / s 2)(44.1 m-4.9 m)

P E =(9.8 kg m / s 2)(39.2 m)
P E =3 8 4 . 1 6 𝐉

The Kinetic Energy at t=1 s is,

KE =(1)/(2)mv^2

K E =(1)/(2)(1 kg)(9.8 m / s)^2
KE =4 8 . 0 2J

The Total Mechanical Energy is:

MET=PE+KE

M ET=384.16 J+48.02 J
MET=4 3 2 . 1 8J

Summarizing the answers in Table 1, you can see clearly the equivalence of the Total Mechanical Energy in every second.

Following the steps in getting the Kinetic Energy and Potential Energy for t=0 s and t=1 s, complete the table when time (t)=2 s, and time (t)=3 s.
Table 1. Summary of the Mechanical Energy of a Free Falling Body

Time, t
( 𝐬 ) Height,
𝐡
( 𝐦 ) Velocity,
𝐯

(𝐦 / 𝐬) Potential
Energy, PE

(𝐉) Kinetic
Energy, KE

(𝐉) Total Mechanical
Energy,
𝐌 𝐄T=𝐏 𝐄 + 𝐊 𝐄
( 𝐉 )

0 44.1 0 432.18 0 432.18

1 39.2 9.8 384.16 48.02 432.18

2

3

10

Added by Krina E.

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