Question

The figure shows a container of mass m2 = 2.6 kg connected to a block of mass m2 by a cord looped around a frictionless pulley. The cord and pulley have negligible mass. When the container is released from rest, it accelerates at 0.88 m/s2 across the horizontal frictionless surface. What are (a) the tension in the cord and (b) mass m2? 

Answer 1

Solution :

m₁ = 2.6 kg

a = 0.88 m/s²

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Free-body diagrams :

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For block of mass m₁: F_net = T = m₁ a

∴ T = m₁ a = (2.6 kg)(0.88 m/s²) = 2.288 N

Therefore, Tension in the cord will be : T = 2.288 N

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For block of mass m₂: F_net = m₂ g - T = m₂ a

∴ m₂ g - m₂ a = T

∴ m₂ (g - a) = T

∴ m₂ (9.81 m/s² - 0.88 m/s²) = (2.288 N)

∴ m₂ = 0.256 kg

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