The table shows information about four objects resting at the top of a hill.

Four Objects at Rest on a Hill

\begin{tabular}{|l|l|l|}
\hline
Object & Mass [tex]$(kg)$[/tex] & Potential energy [tex]$(J)$[/tex] \\
\hline
W & 50 & 980 \\
\hline
X & 35 & 1,029 \\
\hline
Y & 62 & 1,519 \\
\hline
Z & 24 & 1,176 \\
\hline
\end{tabular}

Which object is on the tallest hill?

A. W
B. X
C. Y
D. Z



Answer :

To determine which object is on the tallest hill, we need to calculate the height of each hill using the potential energy formula:

[tex]\[ E_p = m \cdot g \cdot h \][/tex]

where [tex]\(E_p\)[/tex] is the potential energy, [tex]\(m\)[/tex] is the mass, [tex]\(g\)[/tex] is the acceleration due to gravity (9.8 m/s²), and [tex]\(h\)[/tex] is the height. We can rearrange this formula to solve for height [tex]\(h\)[/tex]:

[tex]\[ h = \frac{E_p}{m \cdot g} \][/tex]

Let's calculate the height for each object step by step.

1. Object W:
[tex]\[ m = 50 \, \text{kg}, \, E_p = 980 \, \text{J} \][/tex]
[tex]\[ h = \frac{980}{50 \times 9.8} = \frac{980}{490} \approx 2 \, \text{m} \][/tex]

2. Object X:
[tex]\[ m = 35 \, \text{kg}, \, E_p = 1{,}029 \, \text{J} \][/tex]
[tex]\[ h = \frac{1{,}029}{35 \times 9.8} = \frac{1{,}029}{343} \approx 3 \, \text{m} \][/tex]

3. Object Y:
[tex]\[ m = 62 \, \text{kg}, \, E_p = 1{,}519 \, \text{J} \][/tex]
[tex]\[ h = \frac{1{,}519}{62 \times 9.8} = \frac{1{,}519}{607.6} \approx 2.5 \, \text{m} \][/tex]

4. Object Z:
[tex]\[ m = 24 \, \text{kg}, \, E_p = 1{,}176 \, \text{J} \][/tex]
[tex]\[ h = \frac{1{,}176}{24 \times 9.8} = \frac{1{,}176}{235.2} \approx 5 \, \text{m} \][/tex]

The calculated heights are:
- Object W: 2 meters
- Object X: 3 meters
- Object Y: 2.5 meters
- Object Z: 5 meters

From these calculations, we see that Object Z is on the tallest hill with a height of approximately 5 meters.

Thus, the answer is:

Z