based on this graph, which chart represents the energy for point a?
height (m) vs. time (s)

first chart: two bars labeled e_p (2) and e_k (4)
second chart: one bar labeled e_p (4), e_k has no bar
third chart: two bars labeled e_p (4) and e_k (2)

Question

based on this graph, which chart represents the energy for point a?
height (m) vs. time (s)

first chart: two bars labeled e_p (2) and e_k (4)
second chart: one bar labeled e_p (4), e_k has no bar
third chart: two bars labeled e_p (4) and e_k (2)

Accepted Answer

# Brief Explanations:
To determine the energy chart for point A, we analyze the height - time graph. Point A has a height (from the graph) that is non - zero. Gravitational potential energy $E_p=mgh$ (where $m$ is mass, $g$ is acceleration due to gravity, and $h$ is height) and kinetic energy $E_k$. At a non - zero height (not at the lowest point or the highest point of the motion shown), there is both potential and kinetic energy. But we also consider the total mechanical energy. Looking at the height at point A (around 4 m? Wait, no, the height at point A is 4 m? Wait, the height - time graph: at time $t = 3$ s, point A has height $h = 4$ m. Now, when an object is in motion (like in a projectile - like or free - fall - like motion with some symmetry here), at a given height, the potential energy depends on height and kinetic energy depends on speed. But also, we can think about the total energy. Wait, in the first peak (at $t = 1$ s), height is 8 m. At $t = 2$ s, height is 0, so all energy is kinetic there. At point A ($t = 3$ s, height = 4 m), the potential energy should be half of the potential energy at the first peak (since $h$ is half, $E_p\propto h$) if mass and $g$ are constant. The first peak had $h = 8$ m, so at $h = 4$ m, $E_p$ is half of the $E_p$ at $h = 8$ m. But at $h = 0$ ( $t = 2$ s), $E_p = 0$ and $E_k$ is maximum (equal to the total energy at $h = 8$ m). At $h = 8$ m, $E_k=0$ and $E_p$ is maximum (total energy). So at $h = 4$ m (point A), $E_p$ is half of the maximum $E_p$, and $E_k$ is also half of the maximum $E_k$ (since total energy $E = E_p+E_k$ is conserved, assuming no air resistance). Wait, but looking at the bar charts:

First, let's recall: $E_p=mgh$, so $E_p$ is proportional to $h$. At $h = 8$ m (first peak), $E_p$ is maximum, $E_k = 0$. At $h = 0$ ( $t = 2$ s), $E_p = 0$, $E_k$ is maximum (equal to the total energy, which is equal to the maximum $E_p$). At point A, $h = 4$ m, so $E_p=\frac{1}{2}\times$ maximum $E_p$, and $E_k=\frac{1}{2}\times$ maximum $E_k$ (since $E = E_p + E_k=$ constant). But the bar charts:

- First chart: $E_p = 2$, $E_k = 4$ → total 6.
- Second chart: $E_p = 4$, $E_k = 0$ → total 4.
- Third chart: $E_p = 4$, $E_k = 2$ → total 6. Wait, no, maybe I messed up. Wait, at $t = 2$ s, height is 0, so $E_p = 0$ and $E_k$ is equal to the total energy (which is equal to the $E_p$ at $t = 1$ s where $h = 8$ m). So total energy $E = E_{p1}=mg\times8$. At point A, $h = 4$ m, so $E_p=mg\times4=\frac{1}{2}E_{p1}$, so $E_p=\frac{1}{2}E$ (since $E = E_{p1}$ at $t = 1$ s). Then $E_k=E - E_p=\frac{1}{2}E$. So $E_p$ and $E_k$ should be equal? Wait, no, that's only if $E_p=\frac{1}{2}E$, then $E_k=\frac{1}{2}E$. But the bar charts:

Wait, the third chart has $E_p = 4$ and $E_k = 2$? No, wait, maybe the height at point A is 4 m, and the first peak is 8 m. So $E_p$ at point A is half of $E_p$ at first peak. At first peak, $E_p$ is maximum (let's say 4 units? No, maybe the bar lengths:

Wait, the first bar chart: $E_p = 2$, $E_k = 4$ → sum 6.

Second: $E_p = 4$, $E_k = 0$ → sum 4.

Third: $E_p = 4$, $E_k = 2$ → sum 6.

Wait, at $t = 2$ s, height is 0, so $E_p = 0$ and $E_k$ is equal to the total energy (which is the same as the $E_p$ at $t = 1$ s, where $h = 8$ m). So total energy $E = E_{p}(t = 1)=mg\times8$. At point A ($t = 3$ s, $h = 4$ m), $E_p=mg\times4=\frac{1}{2}E$, so $E_p=\frac{1}{2}E$, and $E_k=E - E_p=\frac{1}{2}E$. But the bar charts:

Wait, maybe the height at point A is 4 m, so $E_p$ is 4 (proportional to 4) and $E_k$ is 2? No, that doesn't fit. Wait, maybe I made a mistake. Let's look at the time - height graph again. The motion: from $t = 0$ to $t = 1$ s, it goes up to $h = 8$ m (so decelerating, $E_p$ increasing, $E_k$ decreasing). From $t = 1$ to $t = 2$ s, it comes down to $h = 0$ (accelerating, $E_p$ decreasing, $E_k$ increasing). From $t = 2$ to $t = 3$ s, it goes up to $h = 4$ m (decelerating, $E_p$ increasing, $E_k$ decreasing). From $t = 3$ to $t = 4$ s, it comes down to $h = 0$ (accelerating, $E_p$ decreasing, $E_k$ increasing).

At $t = 2$ s, $h = 0$, so $E_p = 0$, $E_k$ is maximum (equal to the total energy, which is $E_p$ at $t = 1$ s: $mg\times8$).

At $t = 3$ s (point A), $h = 4$ m, so $E_p=mg\times4$, which is half of $mg\times8$, so $E_p=\frac{1}{2}\times$ total energy. Then $E_k=$ total energy $- E_p=\frac{1}{2}\times$ total energy. But the bar charts:

Wait, the third chart has $E_p = 4$ and $E_k = 2$. Wait, maybe the total energy at $t = 2$ s ( $h = 0$) is $E_k = 6$ (if we add the first chart's $E_p + E_k=2 + 4 = 6$). Then at point A, $E_p$ should be 3 and $E_k = 3$, but that's not matching. Wait, maybe I misread the height. Wait, the height at point A is 4 m, and the first peak is 8 m. So the ratio of heights is $4/8 = 1/2$. So $E_p$ at A is half of $E_p$ at the first peak. At the first peak, $E_k = 0$, $E_p$ is maximum (let's say 8 units). At $t = 2$ s, $E_p = 0$, $E_k = 8$ units. At point A ( $h = 4$ m), $E_p = 4$ units (half of 8), so $E_k=8 - 4 = 4$ units? No, that can't be. Wait, no, maybe the motion is such that at $t = 2$ s, it's at the bottom, so all energy is kinetic. Then at point A ( $h = 4$ m), which is halfway up from the bottom ( $h = 0$) to the second peak (but the second peak is $h = 4$ m, not 8 m). Wait, the first peak is 8 m, then it goes down to 0, then up to 4 m (point A), then down to 0. So the second upward motion (from $t = 2$ to $t = 3$ s) has a peak of 4 m, which is half of the first peak's 8 m. So the total energy for the second motion: when it goes up to 4 m, the maximum potential energy there is $mg\times4$, so the total energy for the second motion is $mg\times4$ (at the peak of the second motion, $E_k = 0$, $E_p=mg\times4$). But at $t = 2$ s, it has kinetic energy equal to $mg\times4$ (since it's moving up to 4 m). Wait, this is getting confusing. Let's look at the bar charts:

The third chart has $E_p = 4$ and $E_k = 2$. Wait, no, the correct chart should have $E_p$ and $E_k$ such that at point A, there is both potential and kinetic energy, and the sum is equal to the total energy. But also, when the height is 4 m, the potential energy is less than the maximum potential energy (at 8 m) and kinetic energy is less than the maximum kinetic energy (at 0 m). Wait, the first chart: $E_p = 2$, $E_k = 4$ → sum 6. Second: $E_p = 4$, $E_k = 0$ → sum 4. Third: $E_p = 4$, $E_k = 2$ → sum 6.

Wait, at $t = 2$ s, the object is at height 0, so $E_p = 0$ and $E_k$ is equal to the total energy (which is the same as the $E_p$ at $t = 1$ s, where $h = 8$ m). So total energy $E = mg\times8$. At point A ($h = 4$ m), $E_p=mg\times4=\frac{1}{2}E$, so $E_p=\frac{1}{2}E$, and $E_k=E - E_p=\frac{1}{2}E$. So $E_p$ and $E_k$ should be equal? But the bar charts don't have equal bars. Wait, maybe the height at point A is 4 m, and the first peak is 8 m, so the potential energy at A is 4 (proportional to 4) and kinetic energy is 2 (since maybe the total energy is 6? No, this is confusing. Wait, let's look at the bar heights. The third chart has $E_p$ as 4 (taller bar) and $E_k$ as 2 (shorter bar). The first chart has $E_p = 2$ and $E_k = 4$. The second has $E_k = 0$. At point A, the object is moving (since it's going from 4 m to 0 m, so it has speed, so $E_k
eq0$, so second chart is out. Now, between first and third: at $h = 4$ m, which is less than the first peak's 8 m, so $E_p$ should be less than the $E_p$ at the first peak. But at the first peak, $E_k = 0$ and $E_p$ is maximum. At point A, $E_p$ is less than that maximum, and $E_k$ is non - zero. Wait, maybe the key is that at point A, the height is 4 m, so $E_p$ is 4 (from the bar) and $E_k$ is 2, because when it was at 8 m, $E_p$ was 8 (but in the charts, the first peak's $E_p$ isn't shown). Alternatively, think about the motion: when going up to 4 m (point A), it's decelerating, so it has both potential and kinetic energy. The correct chart should have $E_p$ as 4 and $E_k$ as 2 (third chart), because the potential energy is related to height (4 m) and kinetic energy is less than potential energy? Wait, no, when moving up, at a given height, if it's still moving up, speed is decreasing, so kinetic energy is decreasing, but at $h = 4$ m (on the way up to 4 m? No, at $t = 3$ s, it's at the peak of the second motion (height 4 m), so at that instant, is it moving? Wait, at the peak of a motion (like a projectile at the top of its trajectory), the vertical speed is zero. Wait, that's a key point! At the peak of the second motion (point A, $t = 3$ s, height 4 m), the vertical speed is zero, so $E_k = 0$? But that contradicts the earlier thought. Wait, no, the graph: from $t = 2$ to $t = 3$ s, it's going up to 4 m (so it's a rising motion, and at $t = 3$ s, it's at the peak of that small rise, so vertical speed is zero. Then $E_k = 0$ and $E_p = 4$ (since $h = 4$ m). But that would mean the second chart ( $E_p = 4$, $E_k = 0$) is correct. But wait, at $t = 1$ s, it was at the peak of the first motion (8 m), speed zero, $E_k = 0$, $E_p$ maximum. At $t = 2$ s, speed maximum ( $h = 0$), $E_k$ maximum, $E_p = 0$. At $t = 3$ s, it's at the peak of the second motion (4 m), speed zero, so $E_k = 0$, $E_p = 4$ (since $h = 4$ m, which is half of 8 m, so $E_p$ is half of the $E_p$ at $t = 1$ s). So that would mean the second chart ( $E_p = 4$, $E_k = 0$) is correct. But earlier I thought it was moving, but at the peak of a vertical motion, speed is zero. So that's the mistake! At the peak of any vertical motion (like throwing a ball up, at the top, speed is zero), so at point A (the peak of the second small motion), vertical speed is zero, so $E_k = 0$ and $E_p = 4$ (since $h = 4$ m). Therefore, the correct chart is the second one: $E_p = 4$, $E_k = 0$.

# Answer:
B. $ \boldsymbol{E_p = 4, E_k = 0} $ (the second bar chart with $E_p$ as 4 and $E_k$ as 0)

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