Two blocks are connected by massless string that is wrapped around a pulley. Block 1 has a mass m1=6.50 kg, block 2 has a mass m2=2.50kg, while the pulley has a mass of 1.40 kg and a radius of 22.0 cm. When the pulley turns, there is friction in the axel that exerts a torque of magnitude 0.300 N m.

If block 1 is released from rest at a height h=0.900
m, how long does it take to drop to the floor

1 answer

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The acceleration of block 1 can be calculated using Newton's second law:

a = F/m1 = (m2g - 0.300 N m)/6.50 kg = (2.50 kg * 9.81 m/s^2 - 0.300 N m)/6.50 kg = 2.45 m/s^2

The time it takes for block 1 to drop to the floor can be calculated using the equation for motion with constant acceleration:

t = (2*h)/(vf + vi) = (2*0.900 m)/(0 + 0) = 2.00 s