Two blocks connected by a cord passing over a small, frictionless pulley rest on frictionless planes (the figure (figure 1)).
1 answer:
We have that the acceleration when going left is mathematically given as
a = 0.671 m/s^2
<h3> Acceleration of Pulley</h3>
Let the system move leftward in direction
Hence we have tension to be
T - m1gsin53 = m1a
m2gsin30 - T = m2a
Therefore
m2gsin30 - m1gsin53.1 = a(m1+m2)
a=g(m2sin30-m1sin53.1) / (m1+m2)
a = 0.671 m/s^2
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