Given :
A pinball machine launches a .045 kg ball at a speed of 9.2 m/s.
To Find :
The potential energy of the spring just before it launched the pinball.
Solution :
We know, their is no external force applied on system.
It means that kinetic energy will remains constant.
Initial Energy = Final Energy
[tex]K.E_i + P.E_i = K.E_f + P.E_f\\\\0 + P.E_i= \dfrac{mv^2}{2}+ 0\\\\P.E_i = \dfrac{0.045 \times 9.2^2}{2}\ J\\\\P.E_i = 1.9044\ J[/tex]
Therefore, the potential energy of the spring just before it launched the pinball os 1.9044 J.
