Question

Two blocks with masses m1 = 3.8 kg and m2 = 5.6 kg are connected by a string that hangs over a pulley of mass M = 2.1 kg and radius R = 0.2 m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of m2 after it has fallen by 0.7 m. Treat the pulley as a disk.

Answer 1

Answer on Question #56959-Physics-Mechanics-Relativity

Two blocks with masses $m1 = 3.8$ kg and $m2 = 5.6$ kg are connected by a string that hangs over a pulley of mass $M = 2.1$ kg and radius $R = 0.2$ m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of $m2$ after it has fallen by $0.7$ m. Treat the pulley as a disk.

Solution

Using the energy principle

$$m _ {2} g h = m _ {1} g h + \frac {m _ {1} v ^ {2}}{2} + \frac {m _ {2} v ^ {2}}{2} + \frac {I \omega^ {2}}{2}$$ $$(m _ {2} - m _ {1}) g h = \frac {m _ {1} + m _ {2}}{2} v ^ {2} + \frac {I \omega^ {2}}{2}.$$ $$I _ {d i s k} = \frac {M r ^ {2}}{2}.$$

The string does not slip:

$$\omega = \frac {v}{R}.$$

Thus,

$$(m _ {2} - m _ {1}) g h = \frac {m _ {1} + m _ {2}}{2} v ^ {2} + \frac {1}{2} \frac {M R ^ {2}}{2} \left(\frac {v}{R}\right) ^ {2}.$$ $$v = \sqrt {2 g h \frac {m _ {2} - m _ {1}}{m _ {1} + m _ {2} + \frac {M}{2}}} = \sqrt {2 \cdot 9 . 8 \cdot 0 . 7 \frac {5 . 6 - 3 . 8}{3 . 8 + 5 . 6 + \frac {2 . 1}{2}}} = 1. 5 \frac {m}{s}.$$

Answer: $1.5\frac{m}{s}$

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