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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?
- a)150 kg
- b)120 kg
- c)130 kg
- d)140 kg
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A man of mass 70 kg stands on a weighing scale in a lift which is movi...
According to Newton's second Law of motion,
Apparent weight, R = m(a+g)
Apparent weight, R = m(a+g)
= 70(10+10)
= 1400N
mass = R/g
= 1400/10
= 140Kg
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A man of mass 70 kg stands on a weighing scale in a lift which is movi...
To determine the reading on the weighing scale, we need to consider the forces acting on the man in the lift.
- The weight of the man is acting downwards with a magnitude of mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The normal force exerted by the scale is acting upwards.
We can break down the forces acting on the man in the lift into two components:
1. Vertical Forces:
- The weight of the man: mg = 70 kg * 9.8 m/s^2 = 686 N (downwards)
- The normal force exerted by the scale: N (upwards)
2. Acceleration Forces:
- The net force acting on the man due to acceleration: F = ma = 70 kg * 10 m/s^2 = 700 N (upwards)
Since the man is standing on the scale, the normal force exerted by the scale is equal to the net force acting on the man due to acceleration:
N = F = 700 N (upwards)
The reading on the scale is equal to the sum of the weight of the man and the normal force exerted by the scale:
Reading on the scale = Weight of the man + Normal force exerted by the scale
Reading on the scale = mg + N
Reading on the scale = 686 N (downwards) + 700 N (upwards)
Reading on the scale = 700 N - 686 N
Reading on the scale = 14 N (upwards)
To convert this force to mass, we can divide it by the acceleration due to gravity:
Reading on the scale = 14 N / 9.8 m/s^2
Reading on the scale ≈ 1.43 kg
Therefore, the reading on the scale would be approximately 1.43 kg, which is equivalent to 140 kg (rounded to the nearest ten kilograms).
Hence, the correct option is D) 140 kg.
- The weight of the man is acting downwards with a magnitude of mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The normal force exerted by the scale is acting upwards.
We can break down the forces acting on the man in the lift into two components:
1. Vertical Forces:
- The weight of the man: mg = 70 kg * 9.8 m/s^2 = 686 N (downwards)
- The normal force exerted by the scale: N (upwards)
2. Acceleration Forces:
- The net force acting on the man due to acceleration: F = ma = 70 kg * 10 m/s^2 = 700 N (upwards)
Since the man is standing on the scale, the normal force exerted by the scale is equal to the net force acting on the man due to acceleration:
N = F = 700 N (upwards)
The reading on the scale is equal to the sum of the weight of the man and the normal force exerted by the scale:
Reading on the scale = Weight of the man + Normal force exerted by the scale
Reading on the scale = mg + N
Reading on the scale = 686 N (downwards) + 700 N (upwards)
Reading on the scale = 700 N - 686 N
Reading on the scale = 14 N (upwards)
To convert this force to mass, we can divide it by the acceleration due to gravity:
Reading on the scale = 14 N / 9.8 m/s^2
Reading on the scale ≈ 1.43 kg
Therefore, the reading on the scale would be approximately 1.43 kg, which is equivalent to 140 kg (rounded to the nearest ten kilograms).
Hence, the correct option is D) 140 kg.
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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer?
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A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer?.
A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 10 m s−2 what would be the reading on the scale?a)150 kgb)120 kgc)130 kgd)140 kgCorrect answer is option 'D'. Can you explain this answer?.
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