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posted by javane Friday, November 26, 2010 at 1:43am.

A 1.0-kg mass (mA) and a 7.0-kg mass (mB) are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. Find the acceleration of the larger mass

a = [(mB-mA)/(mB+mA)]*g = (3/4)g You can derive it by writing free-body motion equations for each mass, and eliminating the cord tension force, T. mA*a = T - mA*g mB*a = mB*g -T (mB + mA)*a = (mB - mA)*g

9 m/s2

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