Physic Labs

Problem 3

Two masses m1=3.0 kgm_1=3.0\,\mathrm{kg} and m2=5.0 kgm_2=5.0\,\mathrm{kg} hang from a light inextensible string over a massless frictionless pulley; the system starts from rest. Take g=10 m s−2g=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 s3.0\,\mathrm s. (c) How far has each mass moved by then?
T=m1(g+a)=37.5 NT=m_1(g+a)=37.5\,\mathrm N
problems.proof.analysis

For m1m_1, tension exceeds its weight while it accelerates upward, so Newton’s law gives the tension directly.

problems.proof.pitfall. Choose downward for m2m_2 as positive; tension does not cancel in the equation for either individual body.