Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 7 - Kinetic Energy and Work - Problems - Page 171: 11b


Angle between force and displacement $= 118^{\circ}$

Work Step by Step

$Work$ $Done= F.d=F\times d cos\theta$ Work Done is equal to Change in Kinetic Energy which in this case is equal to $=-30J$. And magnitude of displacement is: $d= \sqrt {2^{2}+4^{2}+3^{2}}=\sqrt {29}m$ Putting the values and solving we get: $-30 = 12\times \sqrt {29} cos\theta$ $cos\theta = \frac{-30}{64.6}= -0.464$ $\theta = cos^{-1}(-0.464)$ $\theta=117.64^{\circ} \approx 118^{\circ}$
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