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A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.
- a)18.66 cal
- b)10 cal
- c)8.71 cal
- d)9.33 cal
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A body of mass 2kg is dragged on a horizontal surface with a constant ...
Understanding the Problem
A body with a mass of 2 kg is being dragged on a horizontal surface at a constant speed of 2 m/s. The coefficient of friction between the body and the surface is 0.2. We need to find the heat generated due to friction over a period of 5 seconds.
Calculating the Force of Friction
- The formula for the force of friction (F_friction) is given by:
- F_friction = coefficient of friction * normal force
- For a horizontal surface, the normal force (N) equals the weight of the body:
- N = mass * g (where g is approximately 9.81 m/s²)
- Therefore:
- N = 2 kg * 9.81 m/s² = 19.62 N
- F_friction = 0.2 * 19.62 N = 3.924 N
Calculating Work Done Against Friction
- Work done (W) against friction is given by:
- W = F_friction * distance
- The distance traveled in 5 seconds at 2 m/s:
- Distance = speed * time = 2 m/s * 5 s = 10 m
- Thus, the work done:
- W = 3.924 N * 10 m = 39.24 J
Converting Work to Heat
- The heat generated (Q) is equal to the work done against friction.
- To convert joules to calories, use the conversion factor:
- 1 cal = 4.184 J
- Converting the work done to calories:
- Q = 39.24 J / 4.184 J/cal ≈ 9.38 cal
Final Answer
- Rounding off gives approximately 9.33 cal, which corresponds to option D.
In conclusion, the heat generated due to friction in this scenario is approximately 9.33 calories.
A body with a mass of 2 kg is being dragged on a horizontal surface at a constant speed of 2 m/s. The coefficient of friction between the body and the surface is 0.2. We need to find the heat generated due to friction over a period of 5 seconds.
Calculating the Force of Friction
- The formula for the force of friction (F_friction) is given by:
- F_friction = coefficient of friction * normal force
- For a horizontal surface, the normal force (N) equals the weight of the body:
- N = mass * g (where g is approximately 9.81 m/s²)
- Therefore:
- N = 2 kg * 9.81 m/s² = 19.62 N
- F_friction = 0.2 * 19.62 N = 3.924 N
Calculating Work Done Against Friction
- Work done (W) against friction is given by:
- W = F_friction * distance
- The distance traveled in 5 seconds at 2 m/s:
- Distance = speed * time = 2 m/s * 5 s = 10 m
- Thus, the work done:
- W = 3.924 N * 10 m = 39.24 J
Converting Work to Heat
- The heat generated (Q) is equal to the work done against friction.
- To convert joules to calories, use the conversion factor:
- 1 cal = 4.184 J
- Converting the work done to calories:
- Q = 39.24 J / 4.184 J/cal ≈ 9.38 cal
Final Answer
- Rounding off gives approximately 9.33 cal, which corresponds to option D.
In conclusion, the heat generated due to friction in this scenario is approximately 9.33 calories.
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Community Answer
A body of mass 2kg is dragged on a horizontal surface with a constant ...
The work done against the force of friction
= μR × displacement = 0.2 × 2 × 9.8 × 2 (in one second)
= (0.2 × 2 × 9.8 × 2) × 5 (in 5 second)
= 39. 2J
Heat generated = 39.2/4.2 = 9.33 cal
= μR × displacement = 0.2 × 2 × 9.8 × 2 (in one second)
= (0.2 × 2 × 9.8 × 2) × 5 (in 5 second)
= 39. 2J
Heat generated = 39.2/4.2 = 9.33 cal
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A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. 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 A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. Can you explain this answer?.
A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. 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 A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 2kg is dragged on a horizontal surface with a constant speed of 2 m/s. If the coefficient of friction between the body and the surface is 0.2, then find the heat generated in 5 sec.a)18.66 calb)10 calc)8.71 cald)9.33 calCorrect answer is option 'D'. Can you explain this answer?.
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