Question

a 0.7 kg sphere is placed on top of a coiled spring. the coiled spring has 22 j of elastic potential energy. when the spring is released, the sphere moves upward, as shown in the diagram below. neglecting friction and the mass of the spring, what is the change in height (δh) of the sphere? (a) 154.0 m (b) 0.1428571429 m (c) 31.4285714286 m (d) 0.0318181818 m

Explanation:

Step1: Apply energy - conservation principle

The elastic potential energy of the spring $E_{p}$ is converted into gravitational potential energy $E_{g}$ of the sphere at its maximum height. So $E_{p}=E_{g}$. The formula for gravitational potential energy is $E_{g}=mgh$, where $m$ is the mass of the sphere, $g = 9.8\ m/s^{2}$ is the acceleration due to gravity, and $h=\Delta h$ is the change in height. We know $E_{p}=22\ J$ and $m = 0.7\ kg$.

Step2: Solve for $\Delta h$

From $E_{p}=mgh$, we can express $\Delta h$ as $\Delta h=\frac{E_{p}}{mg}$. Substitute $E_{p}=22\ J$, $m = 0.7\ kg$, and $g = 9.8\ m/s^{2}$ into the formula: $\Delta h=\frac{22}{0.7\times9.8}=\frac{22}{6.86}\approx3.148671429\ m$.

Answer:

B. $3.1428571429\ m$