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### Laws of motion NCERT Solutions Class 11 Physics - Solved Exercise Question 5.13

Question 5.13:
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?
Solution:
(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:
Rmg = ma
Where, ma is the net force acting 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 = 700 / g = 700 / 10 = 70 kg

(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 = 350 / 10 = 35 kg

(c) Mass of the man, m = 70 kg
Acceleration, a = 5 m/s2 upward
Using Newton’s second law of motion, we can write the equation of motion as:
Rmg = 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 + mg = 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.