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\[60\text{ }kg\]. Therefore, mass of the person, \[m=60\text{ }kg\] Final reading is 70 kg. Thus, extra force applied on the person is \[\Delta F=(70-60)\times 98=98N\] Hence upward acceleration of the person is \[a=\frac{\Delta F}{m}=\frac{98}{60}=163\,\,\,m/{{s}^{2}}\]
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