As shown below, two blocks hang from the two sides of a pulley of mass M = 8.8 k

Question

As shown below, two blocks hang from the two sides of a pulley of mass M = 8.8 kg and radius R = 44 cm, and each block is connected to the floor by a spring. Block 1 has a mass m_1 = 3.9 kg, and the spring connecting it to the floor has a spring constant k_1 = 130 N/m. Block 2 has a mass m_2 = 6.1 kg, and the spring connecting it to the floor has a spring constant k_2 = 160 N/m. When the two blocks are at the same height above the ground, both springs are unstretched. Block 1 is then pulled down a distance d = 28 cm below this position and released at rest. Find the speed of Block 1 when the two blocks are again at the same height above the floor.

Explanation / Answer

when blocks are at same height assumin it the reference position

then Gravitational PE = 0

and spring PE = 0

when block 1 is down by 0.28 m ,

GPE1 = - m1 g d = - 3.9 x 9.81 x0.28 = - 10.71 J

G PE2 = m2 g d = 6.1 x 9.81 x 0.28 = 16.76 J

Spring PE = (k1 + k2) (d^2) /2 = (160 + 130) (0.28^2) /2

= 11.37 J

KE = 0

Applying energy conservation,

PEi + KEi = PEf+ KEf

(-10.71 + 16.76 + 11.37) + 0 = 0 + (m1 v^2 /2 + m2 v^2 /2 + (M R^2 / 2)(v/R)^2 /2)

17.42 = (3.9 + 6.1 ) v^2 /2 + 8.8 v^2 / 4

17.42 = 5 v^2 + 2.2 v^2

v = 1.56 m/s

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