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300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s2, work done against friction is [2006]
- a)100 J
- b)zero
- c)1000 J
- d)200 J
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
300 J of work is done in sliding a 2 kg block up an inclined plane of ...
Work done against gravity = mg sinθ × d = 2 × 10 × 10 (dsinθ = 10) = 200J
Actual work done = 300J
Work done against friction = 300 – 200 = 100J
Actual work done = 300J
Work done against friction = 300 – 200 = 100J
Most Upvoted Answer
300 J of work is done in sliding a 2 kg block up an inclined plane of ...
Loss in potential energy = mgh
= 2 x 1 0 x 10
= 200 J.
Gain in kinetic energy = work done = 300 J
∴ ∴ Work done against friction = 300 - 200 = 100 J
= 2 x 1 0 x 10
= 200 J.
Gain in kinetic energy = work done = 300 J
∴ ∴ Work done against friction = 300 - 200 = 100 J
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Community Answer
300 J of work is done in sliding a 2 kg block up an inclined plane of ...
Given,
Work done = 300 J
Mass of the block, m = 2 kg
Height of the inclined plane, h = 10 m
Acceleration due to gravity, g = 10 m/s²
To find: Work done against friction
Formula used:
Work done against friction = force of friction × distance moved against friction
Since the block is moved up the inclined plane, the force of gravity acting on the block can be resolved into two components:
1. Force acting perpendicular to the inclined plane = mgh = 2 kg × 10 m/s² × 10 m = 200 J
2. Force acting parallel to the inclined plane = mg sin θ, where θ is the angle of inclination of the plane. Here, sin θ = h/l, where l is the length of the inclined plane. So, mg sin θ = (2 kg) × (10 m/s²) × (10/√101) ≈ 19.80 N
Since the block is moved up the inclined plane, the direction of force of friction is opposite to the direction of motion of the block. So, work done against friction is negative.
Total work done = work done against gravity + work done against friction
300 J = 200 J + work done against friction
work done against friction = 300 J - 200 J = 100 J
Therefore, the work done against friction is 100 J, which is option (a).
Work done = 300 J
Mass of the block, m = 2 kg
Height of the inclined plane, h = 10 m
Acceleration due to gravity, g = 10 m/s²
To find: Work done against friction
Formula used:
Work done against friction = force of friction × distance moved against friction
Since the block is moved up the inclined plane, the force of gravity acting on the block can be resolved into two components:
1. Force acting perpendicular to the inclined plane = mgh = 2 kg × 10 m/s² × 10 m = 200 J
2. Force acting parallel to the inclined plane = mg sin θ, where θ is the angle of inclination of the plane. Here, sin θ = h/l, where l is the length of the inclined plane. So, mg sin θ = (2 kg) × (10 m/s²) × (10/√101) ≈ 19.80 N
Since the block is moved up the inclined plane, the direction of force of friction is opposite to the direction of motion of the block. So, work done against friction is negative.
Total work done = work done against gravity + work done against friction
300 J = 200 J + work done against friction
work done against friction = 300 J - 200 J = 100 J
Therefore, the work done against friction is 100 J, which is option (a).
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300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s2, work done against friction is [2006]a)100 Jb)zeroc)1000 Jd)200 JCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s2, work done against friction is [2006]a)100 Jb)zeroc)1000 Jd)200 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g = 10 m/s2, work done against friction is [2006]a)100 Jb)zeroc)1000 Jd)200 JCorrect answer is option 'A'. Can you explain this answer?.
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