## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 211: 21

#### Answer

The object's speed when it was released was 4.5 m/s

#### Work Step by Step

We can find the time $t$ to fall to the floor as: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(1.0~m)}{9.80~m/s^2}}$ $t = 0.452~s$ We can find the horizontal speed $v_x$ when the object reached the edge of the table as: $v_x = \frac{x}{t}$ $v_x = \frac{0.30~m}{0.452~s}$ $v_x = 0.664~m/s$ We can find the rate of deceleration as the object slid across the table as: $F_f = ma$ $mg~\mu_k = ma$ $a = g~\mu_k$ $a = (9.80~m/s^2)(0.50)$ $a = 4.90~m/s^2$ We can find the speed when the object left the person's hand as; $v_x^2 = v_{x0}^2+2ax$ $v_{x0}^2 = v_x^2-2ax$ $v_{x0} = \sqrt{v_x^2-2ax}$ $v_{x0} = \sqrt{(0.664~m/s)^2-(2)(-4.90~m/s^2)(2.0~m)}$ $v_{x0} = 4.5~m/s$ The object's speed when it was released was 4.5 m/s.

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