an object with a mass 3.8kg is trapped within a potential well represented by the diagram above. the only force this object experiences is responsible for this potential energy function, and this function is comprised of several straight segments. each tick mark on the x - axis represents 1m of distance. the object comes to rest whenever it is at positions x = ±4.33m. find the speed in meters per second of the object as it crosses the origin. provide at least 2 decimal places

Question

Accepted Answer

# Explanation:
## Step1: Apply the conservation of mechanical energy
The total mechanical energy $E$ of the object is conserved. At $x = \pm4.33m$, the object is at rest, so its kinetic energy $K = 0$ and its total mechanical energy $E=U$. At $x = 0$, the total mechanical energy $E = K+U$, where $K=\frac{1}{2}mv^{2}$ and $U=- 12.0J$.

The total mechanical energy $E=-9.0J$ (from the potential energy at $x=\pm4.33m$).

## Step2: Use the energy - conservation formula
We know that $E = K+U$ at $x = 0$. Substituting $E=-9.0J$ and $U=-12.0J$ into $E = K+U$, we get $K=E - U$.

$$
\begin{align*}
K&=-9.0J-(-12.0J)\
K&=3.0J
\end{align*}
$$

Since $K=\frac{1}{2}mv^{2}$, and $m = 3.8kg$, we can solve for $v$.

$$
\begin{align*}
\frac{1}{2}mv^{2}&=3.0J\
v^{2}&=\frac{2K}{m}\
v^{2}&=\frac{2\times3.0}{3.8}\
v^{2}&=\frac{6.0}{3.8}\approx1.5789\
v&=\sqrt{1.5789}\approx1.26m/s
\end{align*}
$$

# Answer:
$1.26$

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QUESTION IMAGE

Question

Explanation:

Step1: Apply the conservation of mechanical energy

Step2: Use the energy - conservation formula

Answer: