A man weighs 80 kg the stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s. What would be the reading on the scale? 400N

Question

A man weighs 80 kg the stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s. What would be the reading on the scale? 400N  (A)  mg sin θ  (B)  (p2t2/m)  (C) zero (D)  (p2t2/2m)

Tardigrade Admin · Accepted Answer

Mass of man m = 80 kg Acceleration of lift = 5 m/s When lift is moving upwards, the reading of weighing scale will be equal to R. The equation of motion gives 1:5:3 (Kg)1/2 (Kg)-1/2 (Kg)2

Q. A man weighs 80 kg the stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s. What would be the reading on the scale? $400 N$

$m g s in θ$

$\frac{p ^{2} t ^{2}}{m}$

zero

$\frac{p ^{2} t ^{2}}{2 m}$

Mass of man m = 80 kg Acceleration of lift = 5 m/s When lift is moving upwards, the reading of weighing scale will be equal to R. The equation of motion gives $1 : 5 : 3$ $(K g)^{1/2}$ $(K g)^{- 1/2}$ $(K g)^{2}$

Q. A man weighs 80 kg the stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s. What would be the reading on the scale? 400N

mgsinθ

mp2t2​