Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 6 - Section 6.3 - Forces That Vary - Example - Page 102: 6.5


6.6 kJ

Work Step by Step

We know that the work is the area under the force versus distance graph. Thus, we find: $W = \int_{x_1}^{x_2}F(x)dx $ $W = \int_{0}^{10}(\mu_k mg)dx $ $W = \int_{0}^{10}((\mu_0 + ax^2) mg)dx $ $W = mg \int_{0}^{10}(\mu_0 + ax^2)dx $ We take the integral and plug in the known values to obtain: $ W \approx 6.6 \ kJ$
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