Physic Labs

Problem 1

Masses m1=2m_1=2 kg and m2=3m_2=3 kg hang from the ends of a light inextensible string passing over a fixed frictionless pulley of negligible radius. The system is released from rest and the string remains taut. Take g=10g=10 m/s2^2 and choose each mass's direction of motion as positive. (a) Determine the direction and acceleration of each mass. (b) Find the string tension. (c) Find their speed after moving s=0.80s=0.80 m from rest, and check it by energy conservation.
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(m2−m1)gs=12(m1+m2)v2⇒v=2(m2−m1)gsm1+m2=1.79 m/s;a=2.0 m/s2, T=24 N(m_2-m_1)gs=\frac12(m_1+m_2)v^2\Rightarrow v=\sqrt{\frac{2(m_2-m_1)gs}{m_1+m_2}}=1.79\ \mathrm{m/s};\quad a=2.0\ \mathrm{m/s^2},\ T=24\ \mathrm N
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The heavy mass's lost gravitational energy minus the light mass's gain is (m2−m1)gs(m_2-m_1)gs, equal to the two masses' total kinetic energy. This independent check gives the same speed; final results are a=2.0a=2.0 m/s2^2, T=24T=24 N, v=1.79v=1.79 m/s.