Problem 1

Two masses

m_1=2.0\,\mathrm{kg}

and

m_2=3.0\,\mathrm{kg}

hang from a light inextensible string over a massless frictionless pulley; the system starts from rest. Take

g=10\,\mathrm{m\,s^{-2}}

. The string stays taut and does not slip on the pulley; its free rotation makes the tensions in the two segments equal. The masses move only vertically, so the lighter mass rises by the same distance and at the same speed that the heavier mass descends. Consider the motion after release. (a) Find the acceleration of each mass and the string tension. (b) Determine their speeds after

3.0\,\mathrm s

. (c) How far has each mass moved by then?

problems.proof.stepOf

problems.proof.prevStep problems.proof.nextStep

T-m_1g=m_1a,\quad m_2g-T=m_2a

problems.proof.analysis

Write one force balance for each hanging mass; the tension acts upward on both bodies. Adding the equations eliminates this internal tension and gives the acceleration.

problems.proof.pitfall. Choose downward for

m_2

as positive; tension does not cancel in the equation for either individual body.