Initially,the spring is at its natural length | vedclass

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(A) Let $m_1 = 0.5 \ kg$ and $m_2 = 1 \ kg$. The acceleration of the system is $a = \frac{F}{m_1 + m_2} = \frac{1}{0.5 + 1} = \frac{1}{1.5} = \frac{2}

Vedclass · Accepted Answer

$\frac{20}{3} \, cm$

Vedclass · Answer

(A) Let $m_1 = 0.5 \ kg$ and $m_2 = 1 \ kg$. The acceleration of the system is $a = \frac{F}{m_1 + m_2} = \frac{1}{0.5 + 1} = \frac{1}{1.5} = \frac{2}{3} \ m/s^2$.At maximum extension $x$,the relative velocity of the blocks is zero. In the frame of the center of mass,or by using the work-energy theorem for the system,the work done by the external force $F$ equals the change in potential energy of the spring plus the change in kinetic energy of the system.Alternatively,using the pseudo-force approach in the frame of the center of mass,the maximum extension $x$ is given by $x = \frac{2F_{red}}{k}$,where $F_{red} = \frac{m_1 m_2}{m_1 + m_2} a = \frac{m_1 m_2}{m_1 + m_2} \cdot \frac{F}{m_1 + m_2} = \frac{m_1 m_2 F}{(m_1 + m_2)^2}$.Substituting the values: $x = \frac{2 \cdot (0.5 \cdot 1) \cdot 1}{(0.5 + 1)^2 \cdot 20} = \frac{1}{1.5^2 \cdot 10} = \frac{1}{2.25 \cdot 10} = \frac{1}{22.5} = \frac{10}{225} = \frac{2}{45} \ m$.Wait,let's re-evaluate using the work done by the force $F$ on the center of mass: $W = F \cdot x_{cm} = \frac{1}{2} k x^2$. The extension $x$ is $x = \frac{2 F m_2}{k(m_1 + m_2)} = \frac{2 \cdot 1 \cdot 1}{20 \cdot (0.5 + 1)} = \frac{2}{20 \cdot 1.5} = \frac{2}{30} = \frac{1}{15} \ m$.Converting to cm: $\frac{1}{15} 	imes 100 \ cm = \frac{100}{15} \ cm = \frac{20}{3} \ cm$.

Initially,the spring is at its natural length and both blocks are at rest. Determine the maximum extension in the spring. Given $k = 20 \ N/m$,$m_1 = 0.5 \ kg$,$m_2 = 1 \ kg$,and $F = 1 \ N$.

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