Three masses m\(_{1}\) = 200 kg, m\(_{2}\) = 300 kg and m\(_{3}\) = 400 kg are kept at the vertices of on equilateral triangle of side 20 m. If the masses are shifted to new configuration such that they are at the vertices of an equilateral triangle of 25 m now. Find the work done in this process :
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Remember that work done by an external agent against a conservative field (like gravity) increases the potential energy of the system. Since the masses are moved farther apart, the potential energy becomes less negative (i.e., it increases), so the work done by the external agent must be positive. This can be a quick check for your answer's sign.
Step 1: Understanding the Question:
We are asked to find the work done by an external agent to change the configuration of a system of three masses from one equilateral triangle arrangement to another with a different side length. Step 2: Key Formula or Approach:
The work done by an external agent in changing the configuration of a system is equal to the change in its potential energy.
\[ W_{ext} = \Delta U = U_{final} - U_{initial} \]
The gravitational potential energy of a system of three masses is:
\[ U = -G \left( \frac{m_1 m_2}{r_{12}} + \frac{m_2 m_3}{r_{23}} + \frac{m_3 m_1}{r_{31}} \right) \]
For an equilateral triangle, \(r_{12} = r_{23} = r_{31} = r\), where r is the side length. Step 3: Detailed Explanation:
Let's first calculate the sum of the products of masses:
\[ \Sigma m_i m_j = m_1 m_2 + m_2 m_3 + m_3 m_1 \]
\[ \Sigma m_i m_j = (200)(300) + (300)(400) + (400)(200) \]
\[ \Sigma m_i m_j = 60000 + 120000 + 80000 = 260000 = 26 \times 10^4 \text{ kg}^2 \]
The gravitational constant is \(G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2\). Initial Potential Energy (\(U_i\)):
The initial side length is \(r_i = 20\) m.
\[ U_i = -\frac{G}{r_i} (\Sigma m_i m_j) = -\frac{6.67 \times 10^{-11}}{20} (26 \times 10^4) \]
\[ U_i = -(0.3335 \times 10^{-11}) (26 \times 10^4) = -8.671 \times 10^{-7} \text{ J} \]
Final Potential Energy (\(U_f\)):
The final side length is \(r_f = 25\) m.
\[ U_f = -\frac{G}{r_f} (\Sigma m_i m_j) = -\frac{6.67 \times 10^{-11}}{25} (26 \times 10^4) \]
\[ U_f = -(0.2668 \times 10^{-11}) (26 \times 10^4) = -6.9368 \times 10^{-7} \text{ J} \]
Work Done:
\[ W_{ext} = U_f - U_i = (-6.9368 \times 10^{-7}) - (-8.671 \times 10^{-7}) \]
\[ W_{ext} = (8.671 - 6.9368) \times 10^{-7} = 1.7342 \times 10^{-7} \text{ J} \]
This value is approximately \(1.735 \times 10^{-7}\) J. Step 4: Final Answer:
The work done in this process is \(1.735 \times 10^{-7}\)J.
