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

Question

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.

Answer 1

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

$$T=m_1a,$$ $$m_2g-T=m_2a.$$

Let’s substitute $T$ into the second equation and find the acceleration of the system of the two blocks:

$$m_2g-m_1a=m_2a,$$ $$a=\dfrac{m_2g}{m_1+m_2}=\dfrac{5\ N}{3.5\ kg+\dfrac{5\ N}{9.8\ \dfrac{m}{s^2}}}=1.25\ \dfrac{m}{s^2}.$$

Finally, we can find the tension in the cord:

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

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

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=0+\dfrac{1}{2}\cdot1.25\ \dfrac{m}{s^2}\cdot(2\ s)^2=2.5\ m.$$