A) \[25\,\,m/{{s}^{2}}\]
B) \[2.5\,\,m/{{s}^{2}}\]
C) \[5\,\,m/{{s}^{2}}\]
D) \[10\,\,m/{{s}^{2}}\]
Correct Answer: B
Solution :
Maximum bearable tension in the rope \[T=25\times 20\,=250\,N\] From the figure, T - mg = ma or \[a=\frac{T-mg}{m}\] Given, m = 20 kg, \[g=10\,\,m/{{s}^{2}},\] T = 250 N Hence, \[a=\frac{250-20\times 10}{20}\] \[=\frac{50}{20}=2.5\,m/{{s}^{2}}\]You need to login to perform this action.
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