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) * g/m₁ -m₂). B) m₂ -m₁) * g/m₁ + m₂). C) m₁ + m₂) *g/m₂ -m₁). D) m₂ *g/m₁ + m₂). E) g.

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) * g/m₁ -m₂). B) m₂ -m₁) * g/m₁ + m₂). C) m₁ + m₂) *g/m₂ -m₁). D) m₂ *g/m₁ + m₂). E) g.

Quizplus · Accepted Answer

According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this case, the net force acting on the system is the difference between the weight of the two masses: F = m₂g - m₁g Using the formula for acceleration, we have: a = F/(m₁ + m₂) Substituting the value of F, we get: a = (m₂ - m₁)g/(m₁ + m₂) Simplifying the expression, we get: a = (m₂F1F1F1S1 - m₁F1F1F10)/(m₁ + m₂F1F1F1S1) Therefore, the correct answer is B.

Two Masses M1 and M2 with M2 > M1) Are

Two Masses M₁ and M₂ with M₂ > M₁) Are