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IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ?
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IIn the fig shown, the friction coefficient between the block of mass...
Acceleration of the Plank in the Given Situation
Given Information:
- Mass of the block (m1): 1 kg
- Mass of the plank (m2): 2 kg
- Coefficient of friction between the block and the plank (μ1): 0.4
- Coefficient of friction between the plank and the floor (μ2): 0.1
- Force acting on the upper block (F): 10 N
1. Determine the Force of Friction between the Block and the Plank:
- The force of friction between two surfaces can be calculated using the formula: Ffriction = μ * Fnormal
- The normal force acting on the block is equal to its weight, which can be calculated as: Fnormal = m1 * g, where g is the acceleration due to gravity (9.8 m/s^2)
- Substituting the values, we get: Fnormal = 1 kg * 9.8 m/s^2 = 9.8 N
- Therefore, the force of friction between the block and the plank is: Ffriction1 = 0.4 * 9.8 N = 3.92 N
2. Determine the Force of Friction between the Plank and the Floor:
- Similar to the previous step, the force of friction can be calculated as: Ffriction2 = μ2 * Fnormal2
- The normal force acting on the plank is equal to the sum of the weights of the block and the plank: Fnormal2 = (m1 + m2) * g
- Substituting the values, we get: Fnormal2 = (1 kg + 2 kg) * 9.8 m/s^2 = 29.4 N
- Therefore, the force of friction between the plank and the floor is: Ffriction2 = 0.1 * 29.4 N = 2.94 N
3. Determine the Net Force on the Plank:
- The net force acting on the plank can be calculated by subtracting the force of friction between the block and the plank from the applied force: Fnet = F - Ffriction1
- Substituting the values, we get: Fnet = 10 N - 3.92 N = 6.08 N
4. Determine the Acceleration of the Plank:
- According to Newton's second law of motion, the acceleration of an object is given by the equation: a = Fnet / (m1 + m2)
- Substituting the values, we get: a = 6.08 N / (1 kg + 2 kg) = 2.03 m/s^2
Therefore, the acceleration of the plank when a constant force of 10 N is applied horizontally to the upper 1 kg block is 2.03 m/s^2.
Given Information:
- Mass of the block (m1): 1 kg
- Mass of the plank (m2): 2 kg
- Coefficient of friction between the block and the plank (μ1): 0.4
- Coefficient of friction between the plank and the floor (μ2): 0.1
- Force acting on the upper block (F): 10 N
1. Determine the Force of Friction between the Block and the Plank:
- The force of friction between two surfaces can be calculated using the formula: Ffriction = μ * Fnormal
- The normal force acting on the block is equal to its weight, which can be calculated as: Fnormal = m1 * g, where g is the acceleration due to gravity (9.8 m/s^2)
- Substituting the values, we get: Fnormal = 1 kg * 9.8 m/s^2 = 9.8 N
- Therefore, the force of friction between the block and the plank is: Ffriction1 = 0.4 * 9.8 N = 3.92 N
2. Determine the Force of Friction between the Plank and the Floor:
- Similar to the previous step, the force of friction can be calculated as: Ffriction2 = μ2 * Fnormal2
- The normal force acting on the plank is equal to the sum of the weights of the block and the plank: Fnormal2 = (m1 + m2) * g
- Substituting the values, we get: Fnormal2 = (1 kg + 2 kg) * 9.8 m/s^2 = 29.4 N
- Therefore, the force of friction between the plank and the floor is: Ffriction2 = 0.1 * 29.4 N = 2.94 N
3. Determine the Net Force on the Plank:
- The net force acting on the plank can be calculated by subtracting the force of friction between the block and the plank from the applied force: Fnet = F - Ffriction1
- Substituting the values, we get: Fnet = 10 N - 3.92 N = 6.08 N
4. Determine the Acceleration of the Plank:
- According to Newton's second law of motion, the acceleration of an object is given by the equation: a = Fnet / (m1 + m2)
- Substituting the values, we get: a = 6.08 N / (1 kg + 2 kg) = 2.03 m/s^2
Therefore, the acceleration of the plank when a constant force of 10 N is applied horizontally to the upper 1 kg block is 2.03 m/s^2.
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IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ?
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IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ?.
IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ?.
Solutions for IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? in English & in Hindi are available as part of our courses for Class 11.
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IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ?, a detailed solution for IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? has been provided alongside types of IIn the fig shown, the friction coefficient between the block of mass 1kg and the plank of mass 2kg is 0.4 while that between the plank floor is 0.1. A constant force F starts acting horizontally on the upper 1kg block. The acceleration of plank if F =10N is? ? theory, EduRev gives you an
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