University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 4 - Newton's Laws of Motion - Problems - Exercises - Page 126: 4.35

Answer

(a) $v_0 = 4.85~m/s$ (b) $a = 16.2~m/s^2$ (directed up) (c) $\sum F = 1470~N$ Note that $\sum F = F_N - mg$ (d) He applied an average force of 2360 N to the ground.

Work Step by Step

(a) We can find the speed $v_0$ when he left the floor. $v_0^2 = v^2 - 2gy = 0 -2gy$ $v_0 = \sqrt{-(2)(-9.80~m/s^2)(1.2~m)}$ $v_0 = 4.85~m/s$ (b) We can use $v = 4.85~m/s$ as the final velocity for this part of the question. $a = \frac{v-v_0}{t} = \frac{4.85~m/s-0}{0.300~s}$ $a = 16.2~m/s^2$ (directed up) (c) We can use his weight to find his mass $m$. $weight = mg$ $m = \frac{890~N}{9.80~m/s^2} = 90.8~kg$ $\sum F = ma = (90.8~kg)(16.2~m/s^2)$ $\sum F = 1470~N$ Note that $\sum F = F_N - mg$ (d) $\sum F = F_N - mg$ $F_N = 1470~N + 890~N = 2360~N$ He applied an average force of 2360 N to the ground.
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