Question

The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pulley of negligible friction and negligible mass. The block of mass m1 is a distance h1 above the ground, and the block of mass m2 is a distance h2 above the ground. m2 is larger than m1. The two-block system is released from rest. Which of the following claims correctly describes the outcome after the blocks are released from rest but before the block of m2 reaches the ground? Select two answers.

Answer 1

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

The given parameters:

• Mass of block 1 = m1
• Mass of block 2, = m2
• Height of block 1 above the ground, = h1
• Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

• For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
• For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163