Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems: 37

Answer

$k = 1.4 \times 10^5 N/m$.

Work Step by Step

The car’s mechanical energy is conserved. The initial KE + initial PE equals the final KE + final PE. The car comes to a momentary stop at the end, so all of the initial kinetic energy of the car becomes potential energy of the compressed spring. $$\frac{1}{2}mv_i^2 + 0 = 0 +\frac{1}{2}kx^2 $$ $$\frac{1}{2}(1200 kg)(23.6m/s)^2 + 0 = 0 +\frac{1}{2}k(2.2m)^2 $$ Solve for $k = 1.4 \times 10^5 N/m$.
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