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An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg?
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An object of mass M slides down a hill of height H of arbitrary shape ...
The initial potential energy of the object is mgh, which has been used up for work against the force of friction. In returning the body to its initial position the force performs the same work. In addition, imparts to the object the initial potential energy. As a result the total work will be 2 mgh.
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An object of mass M slides down a hill of height H of arbitrary shape ...
Introduction:
In this scenario, we have an object of mass M sliding down a hill of height H. The hill has an arbitrary shape and the object stops after traveling a certain horizontal path due to friction. The friction coefficient is different for different segments of the path, but it is independent of the velocity and direction of motion. We need to determine the work that a force must perform to return the object to its initial position, and the correct answer is 2mg.
Explanation:
Step 1: Analyzing the motion
When the object slides down the hill, it gains potential energy due to its height H and loses it due to friction. Eventually, it comes to a stop at the end of the horizontal path. The work done by friction to bring the object to a stop is equal to the loss in potential energy.
Step 2: Finding the work done by friction
The work done by friction can be calculated using the formula:
Work = Force × Distance × cos(θ)
In this case, the force is the frictional force, the distance is the horizontal path traveled by the object, and θ is the angle between the force and displacement.
Step 3: Identifying the frictional forces
Since the friction coefficient is different for different segments of the path, we can assume that there are two different frictional forces acting on the object. Let's denote them as F1 and F2.
Step 4: Calculating the work done by F1
The work done by F1 can be calculated as W1 = F1 × Distance × cos(θ1), where θ1 is the angle between F1 and the displacement.
Step 5: Calculating the work done by F2
Similarly, the work done by F2 can be calculated as W2 = F2 × Distance × cos(θ2), where θ2 is the angle between F2 and the displacement.
Step 6: Work done by the net force
To find the total work done by the net force, we need to sum up the work done by F1 and F2.
Total work done = W1 + W2
Step 7: Simplifying the equation
Since the friction coefficient is independent of velocity and direction, we can assume that F1 = F2. Therefore, the total work done becomes:
Total work done = 2 × F × Distance × cos(θ)
Step 8: Expressing the work in terms of mass and acceleration due to gravity
Since F = μmg, where μ is the friction coefficient, the total work done can be rewritten as:
Total work done = 2 × μmg × Distance × cos(θ)
Step 9: Final answer
From the equation above, we can see that the total work done to return the object to its initial position is directly proportional to the product of mass (M), acceleration due to gravity (g), and the friction coefficient (μ). Thus, the correct answer is 2mg.
Conclusion:
The work that a force must perform to return the object to its initial position around the same path is 2mg. This can be derived by considering the frictional forces acting on different segments of the path and calculating the work
In this scenario, we have an object of mass M sliding down a hill of height H. The hill has an arbitrary shape and the object stops after traveling a certain horizontal path due to friction. The friction coefficient is different for different segments of the path, but it is independent of the velocity and direction of motion. We need to determine the work that a force must perform to return the object to its initial position, and the correct answer is 2mg.
Explanation:
Step 1: Analyzing the motion
When the object slides down the hill, it gains potential energy due to its height H and loses it due to friction. Eventually, it comes to a stop at the end of the horizontal path. The work done by friction to bring the object to a stop is equal to the loss in potential energy.
Step 2: Finding the work done by friction
The work done by friction can be calculated using the formula:
Work = Force × Distance × cos(θ)
In this case, the force is the frictional force, the distance is the horizontal path traveled by the object, and θ is the angle between the force and displacement.
Step 3: Identifying the frictional forces
Since the friction coefficient is different for different segments of the path, we can assume that there are two different frictional forces acting on the object. Let's denote them as F1 and F2.
Step 4: Calculating the work done by F1
The work done by F1 can be calculated as W1 = F1 × Distance × cos(θ1), where θ1 is the angle between F1 and the displacement.
Step 5: Calculating the work done by F2
Similarly, the work done by F2 can be calculated as W2 = F2 × Distance × cos(θ2), where θ2 is the angle between F2 and the displacement.
Step 6: Work done by the net force
To find the total work done by the net force, we need to sum up the work done by F1 and F2.
Total work done = W1 + W2
Step 7: Simplifying the equation
Since the friction coefficient is independent of velocity and direction, we can assume that F1 = F2. Therefore, the total work done becomes:
Total work done = 2 × F × Distance × cos(θ)
Step 8: Expressing the work in terms of mass and acceleration due to gravity
Since F = μmg, where μ is the friction coefficient, the total work done can be rewritten as:
Total work done = 2 × μmg × Distance × cos(θ)
Step 9: Final answer
From the equation above, we can see that the total work done to return the object to its initial position is directly proportional to the product of mass (M), acceleration due to gravity (g), and the friction coefficient (μ). Thus, the correct answer is 2mg.
Conclusion:
The work that a force must perform to return the object to its initial position around the same path is 2mg. This can be derived by considering the frictional forces acting on different segments of the path and calculating the work
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An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg?
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An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg?.
An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object of mass M slides down a hill of height H of arbitrary shape and after travelling a certain horizontal path stops because of friction the friction Coefficient is different for different segments for the entire path by is independent of the velocity and direction motion the work that a force must perform to return the object to its initial position around the same parties and the correct answer is 2 mg?.
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