Question

a person of mass 50 kg stands on a weighing scale on a lift if the lift is descents with uniform velocity of 3m/s​

Answer 1

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If the lift is descending with a downward acceleration of 9ms-2 what would be the reading of the weighing scale? (g=10ms-2) . apparent wt. of person =R=m(g-a)=50(10-9)=50N=5kg 

Answer 2

Answer:

Given:-

• Mass of the Body (M) = 50 Kg.
• Acceleration due to gravity (g) = 10 m/s².
• Acceleration of the Lift (a) = 3 m/s².

From F.B.D of the Lift.

As the Lift is Moving Down there is an Pseudo Force(F) acting on it in the opposite direction I.e. In Upward direction. (F = Ma).

Now, From Equilibrium of Body.

$\large\star \: {\boxed{\bold{F_{net} = 0}}}$

$\large{\tt \hookrightarrow F_{net} = 0}$

$\large{\tt \hookrightarrow N + ma = mg}$

$\begin{gathered}\bold{Here}\begin{cases}\text{N Denotes Normal Reaction} \\ \text{ma = F Denotes Pseudo Force} \\ \text{mg = W Denotes Weight of Man}\end{cases}\end{gathered}$

Simplifying,

$\large{\tt \hookrightarrow N + ma = mg}$

$\large{\tt \hookrightarrow N = mg - ma}$

$\large{\tt \hookrightarrow N = m(g - a)}$

• Here "N" is the Reading of the Weighing Scale.

Substituting the values,

$\large{\tt \hookrightarrow N = 50 \: Kg \times(10 - 3) \: m/s^2}$

$\large{\tt \hookrightarrow N = 50 \times(10 - 3)}$

$\large{\tt \hookrightarrow N = 50 \times 7}$

$\huge{\boxed{\boxed{\tt N = 350 \: N}}}$

∴ Reading of Weighing scale will be 350 Newton's.