﻿ A man of mass 70 kg stands on a weighing scale in a lift which is moving, (a) upwards with a uniform speed of 10 m s–1, (b) downwards with a uniform acceleration of 5 m s–2, (c) upwards with a uniform acceleration of 5 m s–2. What would be the readings on the scale in each case?(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? from Physics Laws of Motion Class 11 Manipur Board

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A man of mass 70 kg stands on a weighing scale in a lift which is moving,
(a) upwards with a uniform speed of 10 m s–1

(b) downwards with a uniform acceleration of 5 m s–2

(c) upwards with a uniform acceleration of 5 m s–2.

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

(a)

Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as,

R – mg = ma

where,

ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

R = m

= 70 × 10

= 700 N

Therefore, reading on the weighing scale =

=

=

(b)

Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 , downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + mg = ma

R = m(g – a

= 70 (10 – 5)

= 70 × 5

= 350 N

∴ Reading on the weighing scale = 350 g = = 35 kg

(c)

Mass of the man, m = 70 kg

Acceleration, a = 5 m/supward

Using Newton’s second law of motion, we can write the equation of motion as:

R
– mg = ma

R = m(g + a

= 70 (10 + 5)

= 70 × 15

= 1050 N
Therefore,

Reading on the weighing scale =

=

= 105 kg

(d) When the lift moves freely under gravity, acceleration a =

Using Newton’s second law of motion, we can write the equation of motion as,

R + mg = ma

R = m(g – a

= m(g – g)

= 0

∴ Reading on the weighing scale = = 0 kg

The man will be in a state of weightlessness.
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