A) 16
B) 6
C) 4
D) 8
Correct Answer: C
Solution :
The free body diagram of the person can be drawn as Let the person moves up with an acceleration a, then \[T-60g=60a\] \[\Rightarrow \,\,{{a}_{\max }}=\frac{{{T}_{\max }}=-60g}{60}\] \[=\frac{360-60g}{60}=\frac{360-600}{60}=-ve\] which means it is not possible to climb up on the rope. Even in this problem it is not possible to remain at rest on rope. No option is right. But if they will ask for the acceleration of climbing down, then \[60g-T=60a\] \[\Rightarrow 60g-{{T}_{\max }}=60{{a}_{\min }}\] \[\therefore \] \[{{a}_{\min }}=\frac{60g-360}{60}=4\,m/{{s}^{2}}\]You need to login to perform this action.
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