Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 4 - Dynamics: Newton's Laws of Motion - General Problems - Page 107: 90

Answer

(a) The mass of sand added to the bucket is 10.6 kg. (b) $a = 0.88~m/s^2$

Work Step by Step

(a) The system of the block and the bucket will start moving when the weight of the bucket and sand ($m_s ~g$) is greater than the force of static friction acting on the block. Let's find the mass $m_s$ such that $m_s ~g = F_f$. Let $m_b$ be the mass of the block. $m_s ~g = F_f$ $m_s ~g = m_b ~g ~\mu_s$ $m_s = m_b ~\mu_s = (28.0 ~kg)(0.45) = 12.6 ~kg$ Since the mass of the bucket is 2.00 kg, the mass of sand added to the bucket is 10.6 kg. (b) We can set up a force equation for the system of the block plus the bucket and the sand. The total mass of the system is $m$ = 28.0 kg + 12.6 kg = 40.6 kg: $ma = \sum F$ $ma = m_s ~g - m_b ~g ~\mu_k$ $a = \frac{m_s ~g - m_b ~g ~\mu_k}{m}$ $a = \frac{(12.6 ~kg)(9.80 ~m/s^2) - (28.0 ~kg)(9.80 ~m/s^2)(0.32)}{40.6 ~kg}$ $a = 0.88~m/s^2$
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.