A ball with a mass of 256 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 36 (kg)/s^2 and was compressed by 12/4 m when the ball was released. How high will the ball go?

Nov 26, 2016

The maximum altitude of the ball is $65 m$.

Explanation:

This can be solved using energy conservation,

${K}_{i} + {U}_{i} = {K}_{f} + {U}_{f}$

Where $K$ is the object's kinetic energy, given by $K = \frac{1}{2} m {v}^{2}$ and $U$ is the object's potential energy. The equation for potential energy will depend on which type of potential energy is involved. In this case, that is both spring potential energy and gravitational potential energy.

Spring potential energy is given by U_(sp)=1/2k(Δs)^2, where $k$ is the spring constant and Δs is the change in length of the spring from its equilibrium length (usually assigned a value of $0$).

Gravitational potential energy is given by ${U}_{g} = m g h$, where $m$ is the object's mass, $g$ is the free-fall acceleration ($9.8 \frac{m}{s} ^ 2$), and $h$ is the object's height above some reference point.

Initially, the ball is not moving, so it has $0$ kinetic energy. If we define its initial height as $0$ as well, it also has no gravitational potential energy. All of the energy isn the system is stored as spring potential energy in the compressed spring.

When the spring is released, that potential energy is transferred to the ball. As the ball is launched, it possesses kinetic energy, and the higher it travels, the slower it moves, and the kinetic energy is transformed into gravitational potential energy with increasing height, $h$.

When the ball reaches its maximum altitude, all of the kinetic energy has been transformed into gravitational potential energy as the ball momentarily comes to a stop before falling back toward the earth. In other words, all of the spring potential energy stored in the spring initially is converted into gravitational potential energy stored in the ball at its maximum altitude. This gives us:

${U}_{s p} = {U}_{g}$

1/2k(Δs)^2=mgh

Solving for $h$ to find the ball's maximum height:

h=(1/2k(Δs)^2)/(mg)

$h = \frac{\frac{1}{2} \left(36 \frac{k g}{s} ^ 2\right) {\left(3 m\right)}^{2}}{\left(0.256 k g\right) \left(9.8 \frac{m}{s} ^ 2\right)}$

$h = 65 m$

Note that I have converted the mass to $k g$.

Hope that helps!