# A ball with a mass of 160 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 27 (kg)/s^2 and was compressed by 3/8 m when the ball was released. How high will the ball go?

Dec 17, 2016

h≈1m

#### Explanation:

Ignoring air resistance, friction, etc., this problem can be solved with energy conservation.

${E}_{m e c h} = K + U$

Where $K$ is kinetic energy and $U$ is potential energy. Thus, when energy is conserved in a system it should follow that

${E}_{i} = {E}_{f}$

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

From this relationship we can see that if potential energy, for example, were to decrease, kinetic energy would have to increase in order for energy to be conserved. The opposite is also true. The idea is that energy is being transformed between two forms (potential and kinetic) rather than any of it being lost of gained. When we have to include friction in problems, for example, energy is not conserved, because some of it is lost to heat (${E}_{t h}$).

Initially, assuming the spring is compressed to point where it's height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

${U}_{s p} = {U}_{f} + {K}_{f}$

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. If not, it would continue to fly upwards forever!

Because we are concerned with the maximum altitude of the ball, what we're really asking is at what point all of the potential energy from the spring has been transferred into gravitational potential energy, which is given by ${U}_{g} = m g h$. When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy (remember projectile motion?), and therefore we have only gravitational potential energy finally.

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

Spring potential energy, ${U}_{s} p$ is given by 1/2k(Δs)^2.

1/2k(Δs)^2=mgh

Rearranging to solve for $h$,

h=(1/2k(Δs)^2)/(mg)

Using our known values for mass, the spring constant, and the compression of the spring, we can now calculate a value for $h$.

$h = \frac{\frac{1}{2} \left(27 \frac{N}{m}\right) {\left(\frac{3}{8} m\right)}^{2}}{0.160 k g \cdot 9.8 \frac{m}{s} ^ 2}$

$h = 1.2 m$

or h≈1m