Answer to Question #167170 in Mechanics | Relativity for Nea Mae Cabresos

Question #167170

a 3.5-kg block rest on a horizontal top. A cord is attached to this block, the cord passing over a pulley at the edge of the table. At the end of the cortex is hung another block weighing 5 N. Determine the tension in the cord and the distance the blocks move after 2 seconds.


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Expert's answer
2021-02-28T07:36:11-0500

(a) Let's first find the acceleration of the system of two blocks. Applying the Newton's Second Law of Motion we get:


T=m1a,T=m_1a,m2gT=m2a.m_2g-T=m_2a.

Let’s substitute TT into the second equation and find the acceleration of the system of the two blocks:


m2gm1a=m2a,m_2g-m_1a=m_2a,a=m2gm1+m2=5 N3.5 kg+5 N9.8 ms2=1.25 ms2.a=\dfrac{m_2g}{m_1+m_2}=\dfrac{5\ N}{3.5\ kg+\dfrac{5\ N}{9.8\ \dfrac{m}{s^2}}}=1.25\ \dfrac{m}{s^2}.

Finally, we can find the tension in the cord:


T=m1a=3.5 kg1.25 ms2=4.37 N.T=m_1a=3.5\ kg\cdot1.25\ \dfrac{m}{s^2}=4.37\ N.

(b) We can find the distance the blocks move after 2 seconds from the kinematic equation:


d=v0t+12at2,d=v_0t+\dfrac{1}{2}at^2,d=0+121.25 ms2(2 s)2=2.5 m.d=0+\dfrac{1}{2}\cdot1.25\ \dfrac{m}{s^2}\cdot(2\ s)^2=2.5\ m.

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