Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 5 - Newton's Laws of Motion - Problems and Conceptual Exercises - Page 140: 8

Answer

$F=414N$

Work Step by Step

(a) To find the force, use the equation $$F=ma$$ To find acceleration, use the kinematics equation relating acceleration, initial velocity, final velocity, and distance. This is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Substitute known values of $v_o=3.85m/s$, $v_f=0.00m/s$, and $\Delta x=0.750m$ yields an acceleration of $$a=\frac{-(3.85m/s)^2}{2(0.750m)}=-9.88m/s^2$$ Substituting the known values of $a=9.88m/s^2$ and $m=42.0kg$ yields a force of $$F=(9.88m/s^2)(42.0kg)=414N$$ (b) If distance is shortened, acceleration must be increased. Since acceleration is directly proportional to force, the force would have to increase too.
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