problem 9:
you shoot a ball in the air upwards by compressing a spring. initially the ball is at a position of y = 0. then the spring is compressed to ( y_0=-0.1m ). the spring is released and the ball shoots up. the mass of the ball is 200 g and the spring constant is k = 100 n/m. once the ball reaches a height of y = 0.1 m, what is the speed of the ball?

Question

Accepted Answer

# Explanation:
## Step1: Determine the initial and final energies
The initial energy $E_{i}$ is the sum of the elastic potential energy of the spring $\frac{1}{2}ky_{0}^{2}$ (since $y = 0$ is the reference for gravitational potential energy $U_{g}=mgy$, here $U_{g,i}=0$). The final energy $E_{f}$ is the sum of the gravitational potential energy $mgy$ and the kinetic energy $\frac{1}{2}mv^{2}$. By the conservation of mechanical energy $E_{i}=E_{f}$, so $\frac{1}{2}ky_{0}^{2}=\frac{1}{2}mv^{2}+mgy$.

## Step2: Solve for $v$
First, substitute the given values: $k = 100\space N/m$, $y_{0}=- 0.1\space m$, $m = 0.2\space kg$, $y = 0.1\space m$, $g = 9.8\space m/s^{2}$ into the equation $\frac{1}{2}ky_{0}^{2}=\frac{1}{2}mv^{2}+mgy$.

Rearrange the equation for $v$:

$$
\begin{align*}
\frac{1}{2}mv^{2}&=\frac{1}{2}ky_{0}^{2}-mgy\
v^{2}&=\frac{ky_{0}^{2}-2mgy}{m}\
v&=\sqrt{\frac{ky_{0}^{2}-2mgy}{m}}
\end{align*}
$$

Substitute the values:

$$
\begin{align*}
v&=\sqrt{\frac{100\times(-0.1)^{2}-2\times0.2\times9.8\times0.1}{0.2}}\
&=\sqrt{\frac{100\times0.01 - 0.392}{0.2}}\
&=\sqrt{\frac{1 - 0.392}{0.2}}\
&=\sqrt{\frac{0.608}{0.2}}\
&=\sqrt{3.04}\approx1.74\space m/s
\end{align*}
$$

# Answer:
The speed of the ball is approximately $1.74\space m/s$

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QUESTION IMAGE

Question

Explanation:

Step1: Determine the initial and final energies

Step2: Solve for \(v\)

Answer: