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A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)
- a)W/4
- b)3W/4
- c)W/2
- d)W
Correct answer is option 'B'. Can you explain this answer?
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A man slides down on a telegraphic pole with an acceleration equal to ...
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A man slides down on a telegraphic pole with an acceleration equal to ...
Explanation:
Given:
Acceleration of the man down the telegraphic pole = 1/4 * acceleration due to gravity
We know that the acceleration due to gravity is the acceleration experienced by any object falling freely under the influence of gravity. It is denoted by 'g' and its approximate value is 9.8 m/s^2.
Step 1: Calculate the acceleration due to gravity (g)
g = 9.8 m/s^2
Step 2: Calculate the acceleration of the man down the pole (a)
a = 1/4 * g = 1/4 * 9.8 = 2.45 m/s^2
Step 3: Determine the frictional force between the man and the pole (F)
We know that frictional force (F) is given by the equation:
F = m * a
where m is the mass of the object and a is the acceleration.
In this case, the mass of the man is not given, but we can assume that his weight (w) is equal to his mass, as weight is the force experienced by an object due to gravity.
So, we can substitute w for m in the equation.
F = w * a
Step 4: Substitute the value of 'a' in the equation
F = w * 2.45
Step 5: Simplify the equation
F = 2.45w
Thus, the frictional force between the man and the pole is 2.45 times the weight of the man, which can be written as 2.45w.
Therefore, the correct answer is option B) 3W/4.
Given:
Acceleration of the man down the telegraphic pole = 1/4 * acceleration due to gravity
We know that the acceleration due to gravity is the acceleration experienced by any object falling freely under the influence of gravity. It is denoted by 'g' and its approximate value is 9.8 m/s^2.
Step 1: Calculate the acceleration due to gravity (g)
g = 9.8 m/s^2
Step 2: Calculate the acceleration of the man down the pole (a)
a = 1/4 * g = 1/4 * 9.8 = 2.45 m/s^2
Step 3: Determine the frictional force between the man and the pole (F)
We know that frictional force (F) is given by the equation:
F = m * a
where m is the mass of the object and a is the acceleration.
In this case, the mass of the man is not given, but we can assume that his weight (w) is equal to his mass, as weight is the force experienced by an object due to gravity.
So, we can substitute w for m in the equation.
F = w * a
Step 4: Substitute the value of 'a' in the equation
F = w * 2.45
Step 5: Simplify the equation
F = 2.45w
Thus, the frictional force between the man and the pole is 2.45 times the weight of the man, which can be written as 2.45w.
Therefore, the correct answer is option B) 3W/4.
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A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect 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 slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect 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 slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect answer is option 'B'. Can you explain this answer?.
A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect 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 slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect 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 slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and poles is equal to (in terms of man’s weight w)a)W/4 b)3W/4c)W/2 d)WCorrect answer is option 'B'. Can you explain this answer?.
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