Problem — Ball launched by a spring

Compressed spring. A spring with $k = 200\;\mathrm{N/m}$ is compressed by $10\;\mathrm{cm}$. Calculate the stored energy and the speed of a $50\;\mathrm{g}$ ball launched by the spring as it returns to the equilibrium position.

Solution

Data: $k = 200\;\mathrm{N/m}$, $x = 10\;\mathrm{cm} = 0{,}10\;\mathrm{m}$, $m = 50\;\mathrm{g} = 0{,}05\;\mathrm{kg}$.

Stored elastic energy:

$U = \tfrac{1}{2}kx^2 = \tfrac{1}{2}(200)(0{,}10)^2 = \tfrac{1}{2}(200)(0{,}01) = 1{,}0\;\mathrm{J}$

$\ev{U = 1{,}0\;\mathrm{J}}$

Launch speed. When the spring returns to equilibrium, all the elastic energy has become kinetic energy of the ball (neglecting friction):

$U = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{\frac{2U}{m}} = \sqrt{\frac{2\cdot 1{,}0}{0{,}05}} = \sqrt{40} \approx 6{,}3\;\mathrm{m/s}$

$\ev{v \approx 6{,}3\;\mathrm{m/s}}$

Physical reading. The spring acts as an energy reservoir: it absorbs $1\;\mathrm{J}$ during compression and returns it entirely as motion. Since $v \propto 1/\sqrt{m}$, a lighter ball would leave faster for the same energy.

Collegamenti

Argomenti: Work and energy Concetti: Elastic potential energy · Elastic force and Hooke’s law · Conservation of mechanical energy Competenze: Conservation of energy Oggetti: Spring