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The table shows the mechanical energy and velocity of a rock that was thrown four times. The rock has a mass of 2.4 kg.

\begin{tabular}{|c|c|c|}
\hline
Trial & Mechanical Energy (J) & Velocity (m/s) \\
\hline
1 & 176.4 & 7.0 \\
\hline
2 & 157.7 & 7.0 \\
\hline
3 & 123.7 & 6.0 \\
\hline
4 & 123.7 & 3.0 \\
\hline
\end{tabular}

During which trial was the rock the highest above the ground?

A. 1
B. 2
C. 3
D. 4


Sagot :

To determine during which trial the rock was highest above the ground, we need to calculate the height of the rock in each trial. The given table includes the mechanical energy and velocity for each trial, and we have the mass of the rock [tex]\( m = 2.4 \)[/tex] kg.

Step-by-Step Solution:

1. Calculate Potential Energy for Each Trial:

Mechanical energy (ME) is the sum of kinetic energy (KE) and potential energy (PE). We will use the formula:
[tex]\[ PE = ME - KE \][/tex]
The kinetic energy can be calculated using:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Let's calculate the potential energy for each trial:

- Trial 1:
[tex]\[ ME = 176.4 \text{ J}, \quad v = 7.0 \text{ m/s} \][/tex]
[tex]\[ KE = \frac{1}{2} \times 2.4 \text{ kg} \times (7.0 \text{ m/s})^2 = 58.8 \text{ J} \][/tex]
[tex]\[ PE = 176.4 \text{ J} - 58.8 \text{ J} = 117.6 \text{ J} \][/tex]

- Trial 2:
[tex]\[ ME = 1577 \text{ J}, \quad v = 7.0 \text{ m/s} \][/tex]
[tex]\[ KE = \frac{1}{2} \times 2.4 \text{ kg} \times (7.0 \text{ m/s})^2 = 58.8 \text{ J} \][/tex]
[tex]\[ PE = 1577 \text{ J} - 58.8 \text{ J} = 1518.2 \text{ J} \][/tex]

- Trial 3:
[tex]\[ ME = \frac{157.7}{1702} \text{ J}, \quad v = 2.0 \text{ m/s} \][/tex]
[tex]\[ ME = 0.092625 \text{ J} \quad (\text{after calculation}) \][/tex]
[tex]\[ KE = \frac{1}{2} \times 2.4 \text{ kg} \times (2.0 \text{ m/s})^2 = 4.8 \text{ J} \][/tex]
[tex]\[ PE = 0.092625 \text{ J} - 4.8 \text{ J} = -4.7073443 \text{ J} \][/tex]

- Trial 4:
[tex]\[ ME = 123.7 \text{ J}, \quad v = 6.0 \text{ m/s} \][/tex]
[tex]\[ KE = \frac{1}{2} \times 2.4 \text{ kg} \times (6.0 \text{ m/s})^2 = 43.2 \text{ J} \][/tex]
[tex]\[ PE = 123.7 \text{ J} - 43.2 \text{ J} = 80.5 \text{ J} \][/tex]

2. Calculate Height for Each Trial:

Potential energy is given by:
[tex]\[ PE = mgh \][/tex]
Solving for height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{PE}{mg} \][/tex]

Using [tex]\( g = 9.8 \text{ m/s}^2 \)[/tex]:

- Trial 1:
[tex]\[ h = \frac{117.6 \text{ J}}{2.4 \text{ kg} \times 9.8 \text{ m/s}^2} = 5.0 \text{ m} \][/tex]

- Trial 2:
[tex]\[ h = \frac{1518.2 \text{ J}}{2.4 \text{ kg} \times 9.8 \text{ m/s}^2} = 64.549 \text{ m} \][/tex]

- Trial 3:
[tex]\[ h = \frac{-4.7073443 \text{ J}}{2.4 \text{ kg} \times 9.8 \text{ m/s}^2} = -0.2001 \text{ m} \ (\text{negative height}) \][/tex]

- Trial 4:
[tex]\[ h = \frac{80.5 \text{ J}}{2.4 \text{ kg} \times 9.8 \text{ m/s}^2} = 3.423 \text{ m} \][/tex]

3. Determine the Trial with Maximum Height:

Comparing the heights:
- Trial 1: 5.0 m
- Trial 2: 64.549 m
- Trial 3: -0.2001 m (not a valid height)
- Trial 4: 3.423 m

The tallest height the rock achieved was during Trial 2 at 64.549 meters.

Therefore, the rock was highest above the ground during Trial 2.