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 127: 4.43

Answer

(a) Please refer to the free body diagram below. (b) The ground exerts a force of 0.0493 N on the froghopper. (c) The normal force is 409 times the froghopper's weight.
1504799541

Work Step by Step

(a) Please refer to the free body diagram below. (b) We can find the acceleration during the jump. $a = \frac{v-v_0}{t} = \frac{4.0~m/s - 0}{0.0010~s}$ $a = 4000~m/s^2$ We can use a force equation to find $F_N$. $\sum F = ma$ $F_N - mg = ma$ $F_N = ma + mg$ $F_N = (1.23\times 10^{-5}~kg)(4000~m/s^2)+(1.23\times 10^{-5}~kg)(9.80~m/s^2)$ $F_N = 0.0493~N$ The ground exerts a force of 0.0493 N on the froghopper. (c) We can express the force in terms of the froghopper's weight $mg$. $F = \frac{0.0493~N}{(1.23\times 10^{-5}~kg)(9.80~m/s^2)}$ $F = 409\times ~weight$ The normal force is 409 times the froghopper's weight.
Small 1504799541
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.