Question

Two masses are connected on a frictionless pulley. M1 = 9 kg sits on a frictionless horizontal surface and is connected to a spring with a spring constant of k = 250 N/m. M2 = 6kg and hangs off the edge of the surface. Initially, both masses are at rest and the spring is at its equilibrium. After the blocks are released, how fast will they be moving when mass 2 falls .1 m?

Answer 1

The conservation of energy for masses is

$\frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 = -m_1 g h \sin \theta + m_2 g h + \frac{1}{2}kh^2$

$v = \sqrt{\frac{2 g h \left ( m_2 -m_1 \sin \theta \right ) + kh^2}{m_1+m_2}}$

The angle is not mention in the question. plug in values for the answer

$v = \sqrt{\frac{2 (9.8) (1) \left ( 6 -9 \sin \theta \right ) + (250)(1)^2}{6+9}}$