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Body of mass √2kg is taken up an incline plane of length 10 m.The coefficient of friction between body & incline is 0.5 .Find the work done by frictional force downward and upward?
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Body of mass √2kg is taken up an incline plane of length 10 m.The coef...
Given Data
- Mass of the body (m) = √2 kg
- Length of the incline (L) = 10 m
- Coefficient of friction (μ) = 0.5
Calculating Normal Force
- The normal force (N) acting on the body on the incline is given by:
N = m * g * cos(θ)
- Here, θ is the angle of inclination, and g is the acceleration due to gravity (approximately 9.81 m/s²).
Frictional Force
- The frictional force (f) can be calculated as:
f = μ * N
- Substituting N, we get:
f = μ * m * g * cos(θ)
Work Done by Friction
1. **Work Done Downward:**
- When the body moves down the incline, the direction of friction is opposite to the movement. Thus, the work done by the frictional force (W_down) is:
W_down = -f * L
- Substituting f:
W_down = -μ * m * g * cos(θ) * L
2. **Work Done Upward:**
- When the body moves upward, the frictional force still opposes the direction of motion. Therefore, the work done by friction (W_up) is:
W_up = -f * L
- Again substituting f gives:
W_up = -μ * m * g * cos(θ) * L
Conclusion
- The work done by friction is negative in both scenarios, indicating that friction opposes the motion of the body. The exact values depend on calculating the normal force and the angle of the incline.
- Mass of the body (m) = √2 kg
- Length of the incline (L) = 10 m
- Coefficient of friction (μ) = 0.5
Calculating Normal Force
- The normal force (N) acting on the body on the incline is given by:
N = m * g * cos(θ)
- Here, θ is the angle of inclination, and g is the acceleration due to gravity (approximately 9.81 m/s²).
Frictional Force
- The frictional force (f) can be calculated as:
f = μ * N
- Substituting N, we get:
f = μ * m * g * cos(θ)
Work Done by Friction
1. **Work Done Downward:**
- When the body moves down the incline, the direction of friction is opposite to the movement. Thus, the work done by the frictional force (W_down) is:
W_down = -f * L
- Substituting f:
W_down = -μ * m * g * cos(θ) * L
2. **Work Done Upward:**
- When the body moves upward, the frictional force still opposes the direction of motion. Therefore, the work done by friction (W_up) is:
W_up = -f * L
- Again substituting f gives:
W_up = -μ * m * g * cos(θ) * L
Conclusion
- The work done by friction is negative in both scenarios, indicating that friction opposes the motion of the body. The exact values depend on calculating the normal force and the angle of the incline.
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Body of mass √2kg is taken up an incline plane of length 10 m.The coefficient of friction between body & incline is 0.5 .Find the work done by frictional force downward and upward?
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Body of mass √2kg is taken up an incline plane of length 10 m.The coefficient of friction between body & incline is 0.5 .Find the work done by frictional force downward and upward? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Body of mass √2kg is taken up an incline plane of length 10 m.The coefficient of friction between body & incline is 0.5 .Find the work done by frictional force downward and upward? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Body of mass √2kg is taken up an incline plane of length 10 m.The coefficient of friction between body & incline is 0.5 .Find the work done by frictional force downward and upward?.
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