KinematicsMedium
Question
The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is $\_\_\_\_$ $m/s$. Assume pulley is massless and string is light and inextensible (Take $g = 10\text{ }m/s^{2}$ ).
Options
A.7.74
B.7.20
C.6.55
D.4.50
Solution
Sol. [Key Concept: Atwood machine: $(m_{B} - m_{A})gh = \frac{1}{2}(m_{A} + m_{B})v^{2}$; solve for $v = \sqrt{60} \approx 7.74$ m/s.]
$${W_{G} = \Delta K }{m_{B}gh - m_{A}gh = \frac{1}{2}\left( {\text{ }m}_{A} + m_{B} \right)V^{2} }{(6 - 2) \times 10 \times 6 = \frac{1}{2}(8)v^{2} }{v = \sqrt{60}\text{ }m/s }{\nu = 7.746\text{ }m/s }$$
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