Physics: Principles with Applications (7th Edition)

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

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 108: 96

Answer

The skater's initial velocity was 12 m/s.

Work Step by Step

The force of kinetic friction opposed the motion of the skater and brought the skater to a stop. Let's find the magnitude of acceleration: $ma = F_f$ $ma = mg ~\mu_k$ $a = g ~\mu_k$ $a = (9.8 ~m/s^2)(0.10)$ $a = 0.98 ~m/s^2$ Since the force of kinetic friction opposed the skaters motion, the acceleration is $-0.98 ~m/s^2$. Therefore, $v^2 - v_0^2 = 2ax$ $-v_0^2 = (2)(-0.98 ~m/s^2)(75 ~m)$ $v_0 = \sqrt{(2)(0.98 ~m/s^2)(75 ~m)}$ $v_0 = 12 ~m/s$ The skater's initial velocity was 12 m/s.
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