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Question

Physics Question on laws of motion

A person is standing on a weighing-scale and observes that the reading is 60kg60\,kg . He then suddenly jumps up and observes that reading goes to 70kg70\,kg . Then his maximum upward acceleration is

A

00

B

1.4m/s2 1.4\,m/s^{2}

C

1.63m/s2 1.63\,m/s^{2}

D

9.8m/s2 9.8\,m/s^{2}

Answer

1.63m/s2 1.63\,m/s^{2}

Explanation

Solution

Initial reading of the machine is 60kg60\, kg.
Therefore, mass of the person, m=60kgm=60\, kg Final reading is 70kg70\, kg.
Thus, extra force applied on the person is
ΔF=(7060)×98=98N\Delta F=(70-60) \times 98=98\, N
Hence upward acceleration of the person is
a=ΔFm=9860=1.63m/s2a=\frac{\Delta F}{m}=\frac{98}{60}=1.63\, m / s ^{2}