A) \[1+\frac{a}{g}\]
B) \[1-\frac{a}{g}\]
C) \[1+\frac{2a}{g}\]
D) \[1-\frac{2a}{g}\]
Correct Answer: C
Solution :
Relative acceleration of M relative to m \[=2a=a(-a)\] In lift frame since, M > m, M is moving downward with acceleration a. \[M(g+a)\,-T=Ma\] Or \[T=Mg\] Or \[T-ma\,-mg=ma\] \[T=m(2a+g)\] Comparing \[Mg=m\,(2a+g)\] \[\frac{M}{m}\,=\frac{2a+g}{g}=1+\frac{2a}{g}\]You need to login to perform this action.
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