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 - 12) and time (s) on x - axis (0 - 4), with a blue line forming a v - like shape: starts at (0,12), goes down to (1.5,0), then up to (3,12), then down to (4,4). point a is at (1,3).
three bar charts:
first: ( e_p ) bar height 3, ( e_k ) bar height 9;
second: ( e_p ) bar height 9, ( e_k ) bar height 3;
third: ( e_p ) bar height 6, ( e_k ) bar height 6

Explanation:

1. First, recall the relationship between gravitational potential energy (\(E_p\)) and kinetic energy (\(E_k\)) in a system with mechanical energy conservation (assuming no air resistance, etc.). Gravitational potential energy is given by \(E_p = mgh\), where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height. Kinetic energy is \(E_k=\frac{1}{2}mv^2\).
2. At the initial time (t = 0), height \(h = 12\) m. So initial potential energy \(E_{p0}=mgh_0 = mg\times12\). The total mechanical energy \(E_{total}=E_{p0}+E_{k0}\). At t = 0, the object is at rest (since it starts moving down), so \(E_{k0} = 0\), thus \(E_{total}=12mg\).
3. At point A, height \(h_A = 3\) m. So potential energy at A, \(E_{pA}=mgh_A=mg\times3\). Since mechanical energy is conserved (assuming no non - conservative forces), \(E_{total}=E_{pA}+E_{kA}\). Substituting \(E_{total}=12mg\) and \(E_{pA}=3mg\), we get \(12mg=3mg + E_{kA}\), so \(E_{kA}=9mg\).
4. Now, looking at the bar charts:

• The first chart has \(E_p = 3\) and \(E_k = 9\), which matches our calculation of \(E_{pA}=3mg\) and \(E_{kA}=9mg\) (the actual values depend on \(mg\), but the ratio and relative values match).
• The second chart has \(E_p = 9\) and \(E_k = 3\), which is the reverse of what we calculated.
• The third chart has \(E_p = 6\) and \(E_k = 6\), which does not match our calculation of \(E_{pA}=3mg\) and \(E_{kA}=9mg\).

Answer:

The first bar chart (with \(E_p = 3\) and \(E_k = 9\))