## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 9-12 - Review Exercises - Page 468: 5

#### Answer

So that the top block does not slip, the maximum speed of the blocks is $0.88~m/s$

#### Work Step by Step

Let $m_t$ be the mass of the top block. When the blocks accelerate, the force of static friction provides the force to accelerate the 1.0-kg block. We can find the maximum acceleration of the two blocks so that the top block does not slip: $m_t~g~\mu_s = m_t~a_m$ $a_m = g~\mu_s$ $a_m = (9.80~m/s^2)(0.45)$ $a_m = 4.41~m/s^2$ We can find the angular frequency of the system: $\omega = \sqrt{\frac{k}{m}}$ $\omega = \sqrt{\frac{150~N/m}{6.0~kg}}$ $\omega = 5.0~rad/s$ We can find the amplitude of the system: $a_m = A~\omega^2$ $A = \frac{a_m}{\omega^2}$ $A = \frac{4.41~m/s^2}{(5.0~rad/s)^2}$ $A = 0.176~m$ We can find the maximum speed of the blocks: $v_m = A~\omega$ $v_m = (0.176~m)(5.0~rad/s)$ $v_m = 0.88~m/s$ So that the top block does not slip, the maximum speed of the blocks is $0.88~m/s$

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