# A ball with a mass of 420 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 28 (kg)/s^2 and was compressed by 9/6 m when the ball was released. How high will the ball go?

Dec 30, 2016

The ball will reach a maximum altitude of $\approx 5.1 m$.

#### Explanation:

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

• Find tl;dr to skip the mini lesson in energy conservation if you feel you have a strong enough grasp of the concepts but are having trouble with the derivation.

Conservation of mechanical energy:

$\textcolor{\mathrm{da} r k b l u e}{\Delta {E}_{\text{mechanical}} = \Delta K + \Delta U}$

where $\Delta K$ is the change in kinetic energy and $\Delta U$ is the change in potential energy.

Thus, when energy is conserved in a system it should follow that:

${E}_{\text{final"=E_"initial}}$

And therefore:

$\Delta E = 0$

$\implies \Delta K + \Delta U = 0$

$\implies {K}_{f} - {K}_{i} + {U}_{f} - {U}_{i} = 0$

$\implies \textcolor{\mathrm{da} r k b l u e}{{K}_{i} + {U}_{i} = {K}_{f} + {U}_{f}}$

• 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 and vice versa.

Initially, assuming the spring is compressed flat against the surface below, all of the energy in the system 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.

${\cancel{K}}_{i} + {U}_{\text{spi}} = {K}_{f} + {U}_{f}$

$\implies {U}_{\text{spi}} = {U}_{f} + {K}_{f}$

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.

• As the ball rises it gains altitude and therefore also gains 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. This is the point at which all of the kinetic energy has been transformed into gravitational potential energy.

• 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.

Gravitational potential energy:

$\textcolor{\mathrm{da} r k b l u e}{{U}_{g} = m g h}$.

• When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy, and therefore we ultimately have only gravitational potential energy.

${U}_{\text{spi"=U_"g}} + \cancel{{K}_{f}}$

$\implies \left(1\right) \text{ } \textcolor{\mathrm{da} r k b l u e}{{U}_{s p} = {U}_{g}}$

Spring potential energy:

color(darkblue)(U_"sp"=1/2k(Δs)^2).

where $k$ is the spring constant and $\Delta s$ is the displacement of the spring from equilibrium

Substituting the potential energy equations into equation $\left(1\right)$:

color(darkblue)(1/2k(Δs)^2=mgh)

( tl;dr ) Rearranging to solve for $h$,

color(crimson)(h=(1/2k(Δs)^2)/(mg))

We have the following information:

• $\mapsto m = 420 \text{ g"=0.420" kg}$
• $\mapsto k = 28 {\text{ kg"//"s}}^{2}$
• $\mapsto \Delta s = \frac{9}{6} \text{ m}$
• $\mapsto g = 9.81 {\text{ m"//"s}}^{2}$

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

h=(1/2(28"kg"//"s"^2)(9/6"m")^2)/(0.420"kg"*9.8"m"//"s"^2)

$\implies \textcolor{c r i m s o n}{h \approx 5.1 \text{m}}$