Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 30

Answer

The frog's speed at the top of the ramp is 2.3 m/s

Work Step by Step

We can find the angle of the ramp; $sin(\theta) = \frac{1.0~m}{2.0~m}$ $\theta = arcsin(\frac{1.0~m}{2.0~m})$ $\theta = 30^{\circ}$ The acceleration directed down the incline is $g~sin(\theta)$. This means that; $v^2 = v_0^2+2ad$ $v = \sqrt{v_0^2+2ad}$ $v = \sqrt{(5.0~m/s)^2+(2)(-9.80~m/s^2)~sin(30^{\circ})(2.0~m)}$ $v = 2.3~m/s$ The frog's speed at the top of the ramp is 2.3 m/s
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