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A block of mass 10kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100N is applied on it, then acceleration of the block will be(Take g = 10 m s-2)
- a)10 m s-2
- b)5 m s-2
- c)15 m s-2
- d)0.5 m s-2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A block of mass 10kg is placed on rough horizontal surface whose coeff...
Here, m = 10 kg, g= 10 m s-2, μ = 0.5, F = 100N
Force of friction
f = μN = μMg = (0.5)(10)(10) = 50N
Force that produce acceleration
F' = F - f = 100 N - 50 N = 50 N
a = F'/ m = 50 N / 10 kg = 5 m s-2
Force of friction
f = μN = μMg = (0.5)(10)(10) = 50N
Force that produce acceleration
F' = F - f = 100 N - 50 N = 50 N
a = F'/ m = 50 N / 10 kg = 5 m s-2
Most Upvoted Answer
A block of mass 10kg is placed on rough horizontal surface whose coeff...
Given Data
- Mass of the block (m) = 10 kg
- Coefficient of friction (μ) = 0.5
- Applied force (F) = 100 N
- Acceleration due to gravity (g) = 10 m/s²
Calculating Frictional Force
The maximum frictional force (f_max) can be calculated using the formula:
- f_max = μ * Normal force
- Normal force = m * g = 10 kg * 10 m/s² = 100 N
- f_max = 0.5 * 100 N = 50 N
Net Force Acting on the Block
The net force (F_net) acting on the block can be calculated as:
- F_net = Applied force - Frictional force
- Since the applied force (100 N) is greater than the frictional force (50 N), the frictional force opposing the motion will be 50 N.
- F_net = 100 N - 50 N = 50 N
Calculating Acceleration
Now, we can calculate the acceleration (a) of the block using Newton's second law:
- F_net = m * a
- 50 N = 10 kg * a
- a = 50 N / 10 kg = 5 m/s²
Conclusion
Thus, the acceleration of the block when a horizontal force of 100 N is applied is:
- 5 m/s²
Therefore, the correct answer is option B.
- Mass of the block (m) = 10 kg
- Coefficient of friction (μ) = 0.5
- Applied force (F) = 100 N
- Acceleration due to gravity (g) = 10 m/s²
Calculating Frictional Force
The maximum frictional force (f_max) can be calculated using the formula:
- f_max = μ * Normal force
- Normal force = m * g = 10 kg * 10 m/s² = 100 N
- f_max = 0.5 * 100 N = 50 N
Net Force Acting on the Block
The net force (F_net) acting on the block can be calculated as:
- F_net = Applied force - Frictional force
- Since the applied force (100 N) is greater than the frictional force (50 N), the frictional force opposing the motion will be 50 N.
- F_net = 100 N - 50 N = 50 N
Calculating Acceleration
Now, we can calculate the acceleration (a) of the block using Newton's second law:
- F_net = m * a
- 50 N = 10 kg * a
- a = 50 N / 10 kg = 5 m/s²
Conclusion
Thus, the acceleration of the block when a horizontal force of 100 N is applied is:
- 5 m/s²
Therefore, the correct answer is option B.
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A block of mass 10kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100N is applied on it, then acceleration of the block will be(Take g = 10 m s-2)a)10 m s-2b)5 m s-2c)15m s-2d)0.5 m s-2Correct answer is option 'B'. 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 block of mass 10kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100N is applied on it, then acceleration of the block will be(Take g = 10 m s-2)a)10 m s-2b)5 m s-2c)15m s-2d)0.5 m s-2Correct answer is option 'B'. 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 block of mass 10kg is placed on rough horizontal surface whose coefficient of friction is 0.5. If a horizontal force of 100N is applied on it, then acceleration of the block will be(Take g = 10 m s-2)a)10 m s-2b)5 m s-2c)15m s-2d)0.5 m s-2Correct answer is option 'B'. Can you explain this answer?.
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