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 144: 63

Answer

(a) $99 m/s^2$ (b) $0.20N$ (c) Stay the same. (d) Increase.

Work Step by Step

(a) To find force, use the equation $F=ma$. To find acceleration, use the kinematics equation relating acceleration, distance, initial velocity, and final velocity, which is $$v_f^2=v_o^2+2a\Delta x$$ Solving for $a$ yields $$a=\frac{v_f^2-v_o^2}{2\Delta x}$$ Substituting known values of $v_o=0.0m/s$, $v_f=2.7m/s$, and $\Delta x=3.7cm=0.037m$ yields an acceleration of $$a=\frac{(2.7m/s)^2}{2(0.037m)}=99m/s^2$$ (b) Substituting known values of $a=99m/s^2$ and $m=2.0g=0.0020kg$ into the force equation yields a force of $$F=(0.0020kg)(99m/s^2)=0.20N$$ (c) Acceleration is independent of mass, therefore, as mass increases, acceleration needed stays the same. (d) As mass increases and acceleration remains the same, the force needed must increase. Therefore, as mass increases, force needed must increase as well.
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