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Question -

Aman 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 of5 m s–2,

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

Whatwould be the readings on the scale in each case?

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



Answer -

(a) Mass of the man, m = 70 kg

Acceleration, a = 0

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

R – mg = ma

Where, ma is the net forceacting on the man.

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

∴R = mg

=70 × 10 = 700 N

∴Reading on the weighing scale = 

(b) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 downward

UsingNewton’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 = 

(c) Mass of the man, m = 70 kg

Acceleration, a = 5 m/s2 upward

UsingNewton’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

∴Reading on the weighing scale = 

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

UsingNewton’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 = 

Theman will be in a state of weightlessness.

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