__Question 5.13__:(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?

__Solution__:**(a)**Mass of the man,

*m*= 70 kg

*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.

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