A) \[(2.0\,m/{{s}^{2}})\,i,0\]
B) \[(2.0\,m/{{s}^{2}})\,i,-(2.0\,m/{{s}^{2}})\,i\]
C) \[(6.0\,m/{{s}^{2}})\,i,-(1.0\,m/{{s}^{2}})\,i\]
D) \[(4.0\,m/{{s}^{2}})\,i,0\] (Take \[g=10\,m/{{s}^{2}}\])
Correct Answer: C
Solution :
\[f=0.6\times 10\times 9.8\,N=58.8\,N\] Since the applied force is greater than f therefore the block will be in motion so, we should consider \[{{f}_{k}}\]. \[{{f}_{k}}=0.4\times 10\times 9.8\,N\]or \[{{f}_{k}}=4\times 9.8\,N\] This would cause acceleration of 40 kg block acceleration \[=\frac{4\times 9.8}{40}=(0.98)\,i\,m{{s}^{-2}}=(1)\,i\,m{{s}^{-2}}\] Acceleration of 10 kg block\[\frac{58.8}{10}=(5.88)\,i=(6)\,i\,m/{{s}^{2}}\]
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