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October 20, 2014

October 20, 2014

Posted by **Ramon** on Friday, December 18, 2009 at 6:20pm.

If the spring constant is 54.0 N/m, what was the initial speed of the block? (Ignore energy losses to sound and other factors during the collision.)

- physics -
**drwls**, Friday, December 18, 2009 at 8:18pmThanks for including the dimensions (kg, etc) of the physical quantities this time.

When the spring is fully compressed, the initial kinetic energy (1/2)MV^2 is converted to gravitational energy MgY and spring potential energy (1/2)kY

Y is the vertical distance the spring is compressed (0.045 m)

g is the acceleration of gravity (9.81 m/s^2)

M is the 0.150 kg mass

k is the 54.0 N/m spring constant

(1/2)MV^2 = MgY + (1/2)kY^2

V^2 = 2gY + (k/M)Y^2

Solve for V by plugging in the numbers. The 2gY gravity term may turn out to be negligible compared to (k/M)Y^2

- physics -
**Ramon**, Friday, December 18, 2009 at 8:50pmI just want to thank DRWLS. I sent to Cramster and the answer was wrong several times by them. Again thanks for help.

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