Question

based on this graph, which chart represents the energy for point a?
height (m) vs. time (s)
graph of height (m) on y - axis (0 - 8) and time (s) on x - axis (0 - 4), with a blue line forming a triangle - like shape: from (0,0) up to (1,8), down to (2,0), up to (3,4), and down to (4,0). point a is at (0,2) on the line.
three bar chart options below:
first chart: ( e_p ) bar with height 2, ( e_k ) bar with height 6.
second chart: ( e_p ) bar with height 6, ( e_k ) bar with height 2.
third chart: ( e_p ) bar with height 2, ( e_k ) bar with height 0 (or no bar for ( e_k ))

Explanation:

Step1: Recall Energy Concepts

At point A (height \( h_A \approx 2 \, \text{m} \)), gravitational potential energy \( E_p = mgh \) (proportional to height) and kinetic energy \( E_k \). Initially (at \( t=0 \), height \( 0 \)), \( E_p = 0 \), so all energy is \( E_k \). As height increases, \( E_p \) increases, \( E_k \) decreases (conservation of mechanical energy, assuming no air resistance). At point A, height is low, so \( E_p \) is small, \( E_k \) is large? Wait, no—wait, the first peak is at \( t=1 \), height \( 8 \, \text{m} \). Point A is at \( t \approx 0.5 \), height \( 2 \, \text{m} \). So when moving upward, speed decreases (since height increases, potential energy increases, kinetic energy decreases). Wait, no: when moving up, acceleration is downward (gravity), so speed decreases. So at lower height (point A), speed is higher (kinetic energy higher), potential energy lower. Wait, but let's check the bar graphs. \( E_p \) is proportional to height (since \( E_p = mgh \), mass \( m \) and \( g \) constant, so \( E_p \propto h \)). So at height \( 2 \, \text{m} \), \( E_p \) is proportional to 2, and \( E_k \) would be total energy minus \( E_p \). At the peak (height \( 8 \, \text{m} \)), \( E_k = 0 \) (momentarily at rest), so total energy \( E_{\text{total}} = E_p(\text{peak}) = mg \times 8 \). At point A (height \( 2 \, \text{m} \)), \( E_p = mg \times 2 \), so \( E_k = E_{\text{total}} - E_p = mg \times 8 - mg \times 2 = mg \times 6 \). So \( E_p = 2 \) (proportional), \( E_k = 6 \) (proportional). So the bar graph with \( E_p = 2 \) and \( E_k = 6 \) is the first chart? Wait no, wait the first chart: \( E_p = 2 \), \( E_k = 6 \). Wait, but when moving up, as height increases, \( E_p \) increases, \( E_k \) decreases. So at lower height (point A), \( E_k \) is higher, \( E_p \) lower. So yes, \( E_p = 2 \), \( E_k = 6 \), which is the first chart (leftmost) with \( E_p = 2 \), \( E_k = 6 \). Wait, no—wait the second chart has \( E_p = 6 \), \( E_k = 2 \), which would be at higher height (more \( E_p \), less \( E_k \)). The third chart has \( E_k = 0 \), which is at peak (but peak is at \( 8 \, \text{m} \), not \( 2 \, \text{m} \)). So the correct chart is the first one (left) with \( E_p = 2 \), \( E_k = 6 \).

Step2: Match to Bar Graphs

Check the three charts:

• First chart: \( E_p = 2 \), \( E_k = 6 \) (matches \( E_p \propto 2 \), \( E_k \propto 6 \))
• Second chart: \( E_p = 6 \), \( E_k = 2 \) (would be at higher height, more \( E_p \))
• Third chart: \( E_k = 0 \) (only at peak, not point A)

So the first chart (left) represents the energy at point A.

Answer:

The first bar chart (with \( E_p = 2 \) and \( E_k = 6 \))