(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
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.
1 answer