Q.44. [Figure: Pulley system] The velocity at which 6 kg mass (shown in figure) strikes the ground w

Q: What is the correct answer for this exam Physics Q44?

The correct answer is A. Let $m_A = 2 kg$ and $m_B = 6 kg$. The system is released from rest. The 6 kg mass falls by $h = 6 m$. By conservation of energy, the decrease in potential energy of the system equals the increase in kinetic energy. Decrease in potential energy: $\Delta PE = m_B g h - m_A g h = (m_B - m_A) g h$. Increase in kinetic energy: $\Delta KE = \frac{1}{2} (m_A + m_B) v^2$. So, $(m_B - m_A) g h = \frac{1}{

Q: Q.44. [Figure: Pulley system] The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6m above the ground is _______ m/s. Assume pulley is massless…

The correct answer is A. Let $m_A = 2 kg$ and $m_B = 6 kg$. The system is released from rest. The 6 kg mass falls by $h = 6 m$. By conservation of energy, the decrease in potential energy of the system equals the increase in kinetic energy. Decrease in potential energy: $\Delta PE = m_B g h - m_A g h = (m_B - m_A) g h$. Increase in kinetic energy: $\Delta KE = \frac{1}{2} (m_A + m_B) v^2$. So, $(m_B - m_A) g h = \frac{1}{

Q: What are the options for this this exam Physics question?

A. 74 · B. 20 · C. 55 · D. 50

Question

Q.44. [Figure: Pulley system] The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6m above the ground is _______ m/s. Assume pulley is massless and string is light and inextensible (Take ).

FirstInTest · Accepted Answer

Let $m_A = 2 kg$ and $m_B = 6 kg$. The system is released from rest. The 6 kg mass falls by $h = 6 m$. By conservation of energy, the decrease in potential energy of the system equals the increase in kinetic energy. Decrease in potential energy: $\Delta PE = m_B g h - m_A g h = (m_B - m_A) g h$. Increase in kinetic energy: $\Delta KE = \frac{1}{2} (m_A + m_B) v^2$. So, $(m_B - m_A) g h = \frac{1}{2} (m_A + m_B) v^2$. $(6 - 2) 	imes 10 	imes 6 = \frac{1}{2} (2 + 6) v^2$. $4 	imes 10 	imes 6 = \frac{1}{2} (8) v^2$. $240 = 4 v^2$. $v^2 = 60$. $v = \sqrt{60} m/s$. $v = \sqrt{4 	imes 15} = 2\sqrt{15} m/s$. $2\sqrt{15} \approx 2 	imes 3.87 = 7.74 m/s$. This matches option (1).

Answer

74

Answer

20

Answer

55

Answer

50

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