To extend a spring by 7 cm a job of 14 J is performed. Find the coefficient of elasticity of the system.
PE=kx22→k=2PEx2=2⋅140.072=5714 (N/m)PE=\frac{kx^2}{2}\to k=\frac{2PE}{x^2}=\frac{2\cdot 14}{0.07^2}=5714\ (N/m)PE=2kx2→k=x22PE=0.0722⋅14=5714 (N/m) . Answer
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