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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 5 m s−2 what would be the reading on the scale?
- a)125 kg
- b)105 kg
- c)95 kg Hz
- d)115 kg
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A man of mass 70 kg stands on a weighing scale in a lift which is movi...
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:
R – mg = ma
R = m(g + a)
= 70 (10 + 5) = 70 x 15
= 1050 N
∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg
Most Upvoted Answer
A man of mass 70 kg stands on a weighing scale in a lift which is movi...
To find the reading on the scale, we need to consider the forces acting on the man in the lift.
1. Weight:
The weight of the man is the force exerted by the Earth on him, given by the equation W = mg, where m is the mass of the man and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of the man is W = 70 kg × 9.8 m/s^2 = 686 N.
2. Normal force:
The normal force is the force exerted by the weighing scale on the man. It acts perpendicular to the surface of the scale and has the same magnitude as the weight when the man is at rest or moving with a constant velocity. In this case, however, the lift is moving upwards with an acceleration of 5 m/s^2.
3. Net force:
Since the lift is accelerating upwards, there is a net force acting on the man. This net force is given by the equation F_net = m × a, where m is the mass of the man and a is the acceleration of the lift. Therefore, the net force on the man is F_net = 70 kg × 5 m/s^2 = 350 N.
Now, let's analyze the forces acting on the man when he is in the lift.
- When the lift is at rest or moving with a constant velocity, the normal force is equal to the weight of the man, i.e., 686 N. This is because the man is in equilibrium and the net force on him is zero.
- When the lift is accelerating upwards, the normal force is greater than the weight of the man. This is because the net force on the man is the sum of the weight and the force due to acceleration, which results in a larger normal force.
In this case, the net force is 350 N, which is greater than the weight of the man. Therefore, the normal force exerted by the weighing scale on the man is greater than his weight.
The reading on the scale is equal to the magnitude of the normal force. Hence, the reading on the scale would be 686 N + 350 N = 1036 N.
Converting this force to kilograms, we divide by the acceleration due to gravity (g): 1036 N ÷ 9.8 m/s^2 = 105.7 kg.
Therefore, the reading on the scale would be approximately 105 kg, which corresponds to option B.
1. Weight:
The weight of the man is the force exerted by the Earth on him, given by the equation W = mg, where m is the mass of the man and g is the acceleration due to gravity (approximately 9.8 m/s^2). Therefore, the weight of the man is W = 70 kg × 9.8 m/s^2 = 686 N.
2. Normal force:
The normal force is the force exerted by the weighing scale on the man. It acts perpendicular to the surface of the scale and has the same magnitude as the weight when the man is at rest or moving with a constant velocity. In this case, however, the lift is moving upwards with an acceleration of 5 m/s^2.
3. Net force:
Since the lift is accelerating upwards, there is a net force acting on the man. This net force is given by the equation F_net = m × a, where m is the mass of the man and a is the acceleration of the lift. Therefore, the net force on the man is F_net = 70 kg × 5 m/s^2 = 350 N.
Now, let's analyze the forces acting on the man when he is in the lift.
- When the lift is at rest or moving with a constant velocity, the normal force is equal to the weight of the man, i.e., 686 N. This is because the man is in equilibrium and the net force on him is zero.
- When the lift is accelerating upwards, the normal force is greater than the weight of the man. This is because the net force on the man is the sum of the weight and the force due to acceleration, which results in a larger normal force.
In this case, the net force is 350 N, which is greater than the weight of the man. Therefore, the normal force exerted by the weighing scale on the man is greater than his weight.
The reading on the scale is equal to the magnitude of the normal force. Hence, the reading on the scale would be 686 N + 350 N = 1036 N.
Converting this force to kilograms, we divide by the acceleration due to gravity (g): 1036 N ÷ 9.8 m/s^2 = 105.7 kg.
Therefore, the reading on the scale would be approximately 105 kg, which corresponds to option B.
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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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. 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 5 m s−2 what would be the reading on the scale?a)125 kgb)105 kgc)95 kg Hzd)115 kgCorrect answer is option 'B'. Can you explain this answer?.
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