Physic Labs

Problem 2

Two masses m1=4.0 kgm_1=4.0\,\mathrm{kg} and m2=6.0 kgm_2=6.0\,\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?
a=6−46+4g=2.0 m s−2a=\frac{6-4}{6+4}g=2.0\,\mathrm{m\,s^{-2}}
problems.proof.analysis

Substituting the two masses and gravity gives a positive downward acceleration for m2m_2. This sign agrees with the fact that m2g>m1gm_2g>m_1g.