# Kinetic energy in a spring

I am building a spring launcher for an engineering project and need to calculate spring compression to achieve a projectile velocity of $4$ m/s. The mass $m_p$ of the projectile is $5$ g, the mass of the spring is $7$ g and the spring constant is $1000$ N/m.

Is it right to assume that initial potential energy in the spring = kinetic energy of the marble + kinetic energy of the spring at the point of release?

I can calculate the marble's kinetic energy easily enough using $E_k=\frac12m_pv^2$. However, how would I calculate the spring's kinetic energy? One end is secured, while the other end would be traveling at the same speed as the marble. I think I would have to use integration somehow but not completely sure how to.

• It is not an easy task to include the spring. In many cases you would assume the spring mass negligible to avoid this issue (so that all kinetic energy is given to the object). But otherwise, yes, you must integrate through the length of the spring, since each part of the spring moves at a different speed (moving faster the further away from the fixed end). – Steeven Apr 12 '17 at 14:22

If the mass $m$ of the spring is negligible compared with the mass $M$ of the marble $(m\ll M)$, it is safe to assume that almost all of the energy stored in the spring is converted to KE of the marble.

If the mass $m$ of the real spring is not neglible it can be modelled as an effective mass of $m^*=\frac13m$ on an ideal massless spring with the same spring constant. Then at the launch point the two masses have the same velocity, so the proportion of the energy stored in the compressed spring which is given to the projectile is $\frac{M}{M+m^*}$.