A) zero
B) \[1.4\text{ }m/{{s}^{2}}\]
C) \[1.63\text{ }m/{{s}^{2}}\]
D) \[9.8\text{ }m/{{s}^{2}}\]
Correct Answer: C
Solution :
Initial reading of the machine is 60 kg- Therefore, mass of the person, m = 60 kg Final reading is 70 kg. Thus, extra force applied on the person is \[\Delta F=(70-60)\times 98=98\,N\] Hence, upward acceleration of the person is \[a=\frac{\Delta F}{m}=\frac{98}{60}=1.63\,m/{{s}^{2}}\]
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