question_answer

A load hangs from a travelling crane, moving horizontally with velocity v. If the load is not to swing more than 4 m horizontally, when the crane is stopped suddenly, what is the maximum allowable speed of the crane?

A) 4.05 m/s

B) 4.00 m/s

C) 3.00 m/s

D) 3.50 m/s

Correct Answer: A

Solution :

[a] \[y=10-\sqrt{{{10}^{2}}-{{4}^{2}}}=0.84m\] Thus\[\frac{1}{2}m{{v}^{2}}=mgy\] Or \[v=\sqrt{2gy}=\sqrt{2\times 9.8\times 0.84}\] \[=4.05\text{ }m/s.\]