Posted by **Juan** on Thursday, September 29, 2011 at 9:40pm.

A student places her 442 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by 4.44 cm, then releases the book. What is the book's speed as it slides away? The spring constant is 1102 N/m.

- Physics -
**TucsonGuy**, Wednesday, April 18, 2012 at 8:20pm
(1/2)*K*x^2 = (1/2)*m*v^2

K-Spring constant, 1102 N/m

x-Compression distance, 4.44cm = .0444m

m-Mass of physics book, 442 g = .442 Kg

Re-arrange:

v = x*(K/m)^0.5

v = .0444*(1102/.442)^0.5

v = 2.22 m/s

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