Question #123911

A mass m1=1.8kg is attached to another mass m2=3.2kg with a light inextensible string passing over a frictionless pulley.m2 is surrounded by a viscous liquid .if the acceleration of m2 downward is 0.6m/s^2 . calculate the frictional force acting as a drag on m2

Expert's answer

According to the second Newton's law

"m_1g-T=-m_1a" and "m_2g-T-F_f=m_2a"

We have

"T=m_1g+m_1a"

"m_2g-m_1g-m_1a-F_f=m_2a \\to F_f=m_2g-m_1g-m_1a-m_2a="

"=(m_2-m_1)g-(m_1+m_2)a=(3.2-1.8)\\cdot 9.81-(3.2+1.8)\\cdot 0.6=10.7N"

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