Asked by susanna
A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How fast will the block be moving when it passes through the equilibrium position?
Answers
Answered by
drwls
Potential energy stored in the spring during compression will be (1/2)k(0.1)^2
= 0.5*100*10^-2 = 0.5 J. There will be additional stored gravitational energy of 1 kg*9.8 m/s^2*0.1 m = 1.0 J
When the block passes through the equilibrium position, 1.5 will be converted to kinetic energy, (1/2) m V^2
Solve for V
= 0.5*100*10^-2 = 0.5 J. There will be additional stored gravitational energy of 1 kg*9.8 m/s^2*0.1 m = 1.0 J
When the block passes through the equilibrium position, 1.5 will be converted to kinetic energy, (1/2) m V^2
Solve for V
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