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A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.
(Give answer upto two decimals)
(Give answer upto two decimals)
Correct answer is '0.71'. Can you explain this answer?
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A wooden block having a mass of 1 kg is placed on a table. The block j...
Resolving the forces and finding their horizontal and vertical component
mg = R + Fcosθ
Frictional force = F sinθ
By Eq.(ii) we get
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A wooden block having a mass of 1 kg is placed on a table. The block j...
Given:
- Mass of the block (m) = 1 kg
- Applied force (F) = 10 N
- Angle at which the force is applied (θ) = 45°
- Acceleration due to gravity (g) = 10 m/s²
To find:
- Coefficient of friction between the table and the block
Assumptions:
- The block is at rest initially.
- The block starts moving when the applied force just overcomes the static friction.
Analysis:
1. Resolving the Force:
The applied force can be resolved into two components:
- The vertical component (Fv) = F * sin(θ)
- The horizontal component (Fh) = F * cos(θ)
2. Calculating the Normal Force:
The normal force (N) is the force exerted by the table on the block in the vertical direction. It is equal in magnitude and opposite in direction to the weight of the block.
- Weight of the block (W) = m * g
- Normal force (N) = W = m * g
3. Determining the Frictional Force:
The frictional force (f) opposes the motion of the block and acts parallel to the surface of the table.
- Frictional force (f) = coefficient of friction (μ) * N
4. Equating Forces in the Horizontal Direction:
In order for the block to start moving, the horizontal component of the applied force must be equal to the frictional force.
- Fh = f
5. Solving for the Coefficient of Friction:
Substituting the values into the equation, we get:
- F * cos(θ) = μ * m * g
Rearranging the equation, we find:
- μ = (F * cos(θ)) / (m * g)
Substituting the given values, we get:
- μ = (10 * cos(45°)) / (1 * 10)
- μ = (10 * √2/2) / 10
- μ = √2 / 2
- μ ≈ 0.707
Therefore, the coefficient of friction between the table and the block is approximately 0.707 or 0.7 (rounded to one decimal place).
- Mass of the block (m) = 1 kg
- Applied force (F) = 10 N
- Angle at which the force is applied (θ) = 45°
- Acceleration due to gravity (g) = 10 m/s²
To find:
- Coefficient of friction between the table and the block
Assumptions:
- The block is at rest initially.
- The block starts moving when the applied force just overcomes the static friction.
Analysis:
1. Resolving the Force:
The applied force can be resolved into two components:
- The vertical component (Fv) = F * sin(θ)
- The horizontal component (Fh) = F * cos(θ)
2. Calculating the Normal Force:
The normal force (N) is the force exerted by the table on the block in the vertical direction. It is equal in magnitude and opposite in direction to the weight of the block.
- Weight of the block (W) = m * g
- Normal force (N) = W = m * g
3. Determining the Frictional Force:
The frictional force (f) opposes the motion of the block and acts parallel to the surface of the table.
- Frictional force (f) = coefficient of friction (μ) * N
4. Equating Forces in the Horizontal Direction:
In order for the block to start moving, the horizontal component of the applied force must be equal to the frictional force.
- Fh = f
5. Solving for the Coefficient of Friction:
Substituting the values into the equation, we get:
- F * cos(θ) = μ * m * g
Rearranging the equation, we find:
- μ = (F * cos(θ)) / (m * g)
Substituting the given values, we get:
- μ = (10 * cos(45°)) / (1 * 10)
- μ = (10 * √2/2) / 10
- μ = √2 / 2
- μ ≈ 0.707
Therefore, the coefficient of friction between the table and the block is approximately 0.707 or 0.7 (rounded to one decimal place).
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A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer?
Question Description
A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer?.
A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wooden block having a mass of 1 kg is placed on a table. The block just starts to move, when a force of 10N is applied at 450 to the vertical to push the block. The coefficient of friction between the table and the block, (taking g = 10m/s2) is approximately________.(Give answer upto two decimals)Correct answer is '0.71'. Can you explain this answer?.
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