Page History

{composition-setup}{composition-setup}
{excerpt:hidden=true}Energy and springs.{excerpt}


A certain spring-loaded gun is cocked by compressing its spring by 5.0 cm.  The gun fires a 4.0 g projectile with a speed of 8.0 m/s.  What spring constant is required for the spring?

h4. Solution

{toggle-cloak:id=sys} *System:*  {cloak:id=sys}Projectile as [point particle] plus the gun as a rigid body of infinite mass.{cloak}

{toggle-cloak:id=int} *Interactions:*  {cloak:id=int}The [conservative|conservative force] spring interaction between the gun and the projectile will give rise to [elastic potential energy|Hooke's Law for elastic interactions#epe].{cloak}

{toggle-cloak:id=mod} *Model:* {cloak:id=mod}[Mechanical Energy, External Work, and Internal Non-Conservative Work].{cloak}

{toggle-cloak:id=app} *Approach:*  

{cloak:id=app}

{toggle-cloak:id=diag} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diag}

We will ignore friction and other [non-conservative|non-conservative force] interactions, which means that the [mechanical energy] of the system will be constant.  We will further make the usual assumption that the projectile stops interacting with the spring when the spring returns to its equilibrium position (the projectile has essentially been "fired" at that point).  Thus, appropriate [initial-state final-state diagram] and [energy bar graphs|Diagrams and Mechanical Energy] are:

|!springgun1.png!|!springgun2.png!|
||Initial||Final||

{cloak:diag}

{toggle-cloak:id=math} {color:red} *Mathematical Representation* {color}

{cloak:id=math}

We can now express the fact that the mechanical energy is constant through the Law of Change:

{latex}\begin{large}\[ E_{i} = U_{i} = \frac{1}{2}kx_{i}^{2} = E_{f} = K_{f} = \frac{1}{2}mv_{f}^{2}\]\end{large}{latex}

Solving for _k_ gives:

{latex}\begin{large}\[ k = \frac{mv_{f}^{2}}{x_{i}^{2}} = \mbox{102 N/m}\]\end{large}{latex}

{cloak:math}
{cloak:app}