Two masses m₁ and m₂ (with m₂ m₁)are connected by a massless rope hung over a frictionless light pulley.If the masses are released then the acceleration of m₂ in the downward direction is given by A)(m₂ - m₁)$\times$ g/(m₁ - m₂) B)(m₂ - m₁)$\times$ g/(m₁ + m₂) C)(m₁ + m₂)$\times$g/(m₂ - m₁) D)m₂ $\times$ g/(m₁ + m₂)

Question

Two masses m₁ and m₂ (with m₂ > m₁)are connected by a massless rope hung over a frictionless light pulley.If the masses are released then the acceleration of m₂ in the downward direction is given by A)(m₂ - m₁)$\times$ g/(m₁ - m₂) B)(m₂ - m₁)$\times$ g/(m₁ + m₂) C)(m₁ + m₂)$\times$g/(m₂ - m₁) D)m₂ $\times$ g/(m₁ + m₂)

Quizplus · Accepted Answer

When two masses are connected by a rope and hung on a pulley, the acceleration of the system can be calculated using the formula: a = (m1-m2)g/(m1+m2) where m1 and m2 are the masses of the objects, and g is the acceleration due to gravity. In this problem, m1 is m₁ and m2 is m₂. Since m₂ > m₁, it follows that m1-m2 = - (m₂ - m₁). Therefore, a = (m₂ - m₁)g/(m₁ + m₂) Therefore, the answer is B.

Two Masses M1 and M2 (With M2 > M1)are Connected ×\times×

Two Masses M₁ and M₂ (With M₂ > M₁)are Connected $\times$