Law of conservation of energy:

$\overline{){{\mathbf{K}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{i}}}{\mathbf{+}}{{\mathbf{W}}}_{{\mathbf{nc}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{f}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{f}}}}$ W_{nc} is the work done by non-conservative forces such as friction.

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{mv}}}^{{\mathbf{2}}}}$

The potential energy of a compressed spring:

$\overline{){\mathbf{U}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{k}}{{\mathbf{x}}}^{{\mathbf{2}}}}$

Gravitational potential energy:

$\overline{){\mathbf{U}}{\mathbf{=}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}}$

**(a)**

Let's consider 3 potions of interest:

A student pushes a baseball of *m* = 0.19 kg down onto the top of a vertical spring that has its lower end fixed to a table, compressing the spring a distance of *d* = 0.12 meters. The spring constant of the spring is *k* = 620 N/m. Let the gravitational potential energy be zero at the position of the baseball in the compressed spring.**Randomized Variables**

*m* = 0.19 kg*k* = 620 N/m*d* = 0.12 m

**Part (a) ** The ball is then released. What is its speed, *v*, in meters per second, just after the ball leaves the spring?

**Part (b) ** What is the maximum height, *h*, in meters, that the ball reaches above the equilibrium point?

**Part (c) ** What is the ball’s velocity, in meters per second, at half of the maximum height relative to the equilibrium point?

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