A) 5 m/s
B) 5-5 m/s
C) 5.3 m/s
D) 4.4 m/s
Correct Answer: C
Solution :
At the point A, the energy of the pendulum is entirely PE. At point B, the energy of the pendulum is entirely KE. It means that as the bob pendulum lowers from A to B, PE is converted into KE. Thus, at B, KE = PE. But b% of the PE is dissipated against air resistance. KE at B = 95% of PE at A ... (i) \[\therefore \] From Eq. (i), \[\frac{1}{2}m{{v}^{2}}=\frac{95}{100}\,mgh\] \[{{v}^{2}}=2\times \frac{95}{100}gh=2\times \frac{95}{100}\times 9.8\times 1.5\] \[v=\sqrt{\frac{19\times 9.8\times 1.5}{10}}=\sqrt{27.93}=5.285\,m{{s}^{-1}}\] \[v=5.3\,m{{s}^{-1}}\]You need to login to perform this action.
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