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

Question

A person is standing on a weighing-scale and observes that the reading is 60 kg . He then suddenly jumps up and observes that reading goes to 70 kg . Then his maximum upward acceleration is (A) 0 (B)  1.4 m/s2 (C)  1.63 m/s2 (D)  9.8 m/s2

Tardigrade Admin · Accepted Answer

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 Δ F=(70-60) × 98=98 N Hence upward acceleration of the person is a=(Δ F/m)=(98/60)=1.63 m / s 2

Q. A person is standing on a weighing-scale and observes that the reading is $60 k g$ . He then suddenly jumps up and observes that reading goes to $70 k g$ . Then his maximum upward acceleration is

$0$

$1.4 m / s^{2}$

$1.63 m / s^{2}$

$9.8 m / s^{2}$

Initial reading of the machine is $60 k g$ .
Therefore, mass of the person, $m = 60 k g$ Final reading is $70 k g$ .
Thus, extra force applied on the person is
$Δ F = (70 - 60) \times 98 = 98 N$
Hence upward acceleration of the person is
$a = \frac{Δ F}{m} = \frac{98}{60} = 1.63 m / s^{2}$

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

0

1.4m/s2

1.63m/s2

9.8m/s2

Initial reading of the machine is 60kg. Therefore, mass of the person, m=60kg Final reading is 70kg. Thus, extra force applied on the person is ΔF=(70−60)×98=98N Hence upward acceleration of the person is a=mΔF​=6098​=1.63m/s2

Q. A person is standing on a weighing-scale and observes that the reading is $60 k g$ . He then suddenly jumps up and observes that reading goes to $70 k g$ . Then his maximum upward acceleration is

$0$

$1.4 m / s^{2}$

$1.63 m / s^{2}$

$9.8 m / s^{2}$

Initial reading of the machine is $60 k g$ .
Therefore, mass of the person, $m = 60 k g$ Final reading is $70 k g$ .
Thus, extra force applied on the person is
$Δ F = (70 - 60) \times 98 = 98 N$
Hence upward acceleration of the person is
$a = \frac{Δ F}{m} = \frac{98}{60} = 1.63 m / s^{2}$