### Q.13:- **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?

**Answer:-**(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* – *m*g* = 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*g

= 70 × 10 = 700 N

∴ Reading on the weighing scale = 700 / g = 700 / 10 = 70 kg

(b)** **Mass of the man, *m* = 70 kg

Acceleration, *a* = 5 m/s^{2} downward

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

*R* + *m*g* = ma*

*R* = *m*(g* – a*)

= 70 (10 – 5) = 70 × 5

= 350 N

∴ Reading on the weighing scale = 350 g = 350 / 10 = 35 kg

(c)** **Mass of the man, *m* = 70 kg

Acceleration, *a* = 5 m/s^{2} upward

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

*R* – *m*g* = ma*

*R* = *m*(g* + a*)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg

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

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

*R* + *m*g* = ma*

*R* = *m*(g* – a*)

= *m*(g – g) = 0

∴ Reading on the weighing scale = 0 / g = 0 kg

The man will be in a state of weightlessness.