A vetical spring fixed at its upper end can be elongated by 2 cm under the action of a stretching force of (80g) g. A body of mass 600g is attached to its free end , and then the systemis displaced fromits equilibrium position by 8 cm . Find the energy of the system at this position. The mass is then released. if the energy is conserved , find the velocity of the body 4 cm away from its equilibrium position [g=1000 cm *S^(-2)]

A vetical spring fixed at its upper end can be elongated by 2 cm under

1.	A vetical spring fixed at its upper end can be elongated by 2 cm under the action of a stretching force of (80g) g. A body of mass 600g is attached to its free end , and then the systemis displaced fromits equilibrium position by 8 cm . Find the energy of the system at this position. The mass is then released. if the energy is conserved , find the velocity of the body 4 cm away from its equilibrium position [g=1000 cm *S^(-2)]
Answer» SOLUTION :Spring constant, `k=(80g)/2=40xx1000=40000 dyn cm^(-1)` Elongation of spring due to a force of (80g) g=2 cm `therefore` Extension of the spring due tothe mass of `600 g =(2xx600)/(80) =15 cm`. Potential ENERGY of the spring at this stage `=1/2 kx^2=1/2 40000(15)^2=45xx10^5 erg` Energy required for a further 8 cm elongation `=1/2 xx40000 xx8^2=12.8xx10^5` erg `therefore` Total energy of the SYSTEM `=(45xx10^5+12.8xx10^5)=57.8xx10^5` erg. Again, potential energy for a DISPLACEMENT of 4 cm from the equilibriumposition `=1/2 xx40000xx4^2=3.2xx10^5` erg `therefore` Kinetic energy in this condition `=12.8 xx10^5-3.2xx10^5 =9.6xx 10^5 erg`. If the velocity of the body is v , then `1/2 mv^2 =9.6xx10^5 or, v_2=(2xx9.6xx10^5)/(600)` or, `v=56.568 cms^(-1)`.