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A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]
- a)2.5
- b)10
- c)5
- d)7.5
Correct answer is option 'C'. Can you explain this answer?
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A 100 N force acts horizontally on a block of 10 kg placed on a horizo...
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A 100 N force acts horizontally on a block of 10 kg placed on a horizo...
To solve this problem, we need to use the formula for friction force:
F_friction = friction coefficient * F_normal
where F_normal is the normal force acting on the block, which is equal to the weight of the block:
F_normal = m * g
where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).
In this case, we have:
m = 10 kg
g = 9.8 m/s^2
F_applied = 100 N
friction coefficient = ?
First, we need to find the friction coefficient. We can do this by using the fact that the block is just on the verge of sliding, which means that the friction force is equal to the applied force:
F_friction = F_applied
Plugging in the values, we get:
friction coefficient * F_normal = F_applied
friction coefficient * m * g = F_applied
friction coefficient = F_applied / (m * g)
friction coefficient = 100 N / (10 kg * 9.8 m/s^2)
friction coefficient = 1.02
So the coefficient of friction is 1.02.
Next, we can use the formula for friction force to find the actual friction force acting on the block:
F_friction = friction coefficient * F_normal
F_friction = 1.02 * m * g
F_friction = 1.02 * 10 kg * 9.8 m/s^2
F_friction = 100.4 N
So the friction force acting on the block is 100.4 N.
Finally, we can use Newton's second law (F = ma) to find the acceleration of the block:
F_net = F_applied - F_friction
F_net = 100 N - 100.4 N
F_net = -0.4 N (since the friction force is acting in the opposite direction to the applied force)
ma = -0.4 N (since the acceleration is in the opposite direction to the applied force)
a = -0.4 N / 10 kg
a = -0.04 m/s^2
So the block is decelerating at a rate of 0.04 m/s^2.
F_friction = friction coefficient * F_normal
where F_normal is the normal force acting on the block, which is equal to the weight of the block:
F_normal = m * g
where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).
In this case, we have:
m = 10 kg
g = 9.8 m/s^2
F_applied = 100 N
friction coefficient = ?
First, we need to find the friction coefficient. We can do this by using the fact that the block is just on the verge of sliding, which means that the friction force is equal to the applied force:
F_friction = F_applied
Plugging in the values, we get:
friction coefficient * F_normal = F_applied
friction coefficient * m * g = F_applied
friction coefficient = F_applied / (m * g)
friction coefficient = 100 N / (10 kg * 9.8 m/s^2)
friction coefficient = 1.02
So the coefficient of friction is 1.02.
Next, we can use the formula for friction force to find the actual friction force acting on the block:
F_friction = friction coefficient * F_normal
F_friction = 1.02 * m * g
F_friction = 1.02 * 10 kg * 9.8 m/s^2
F_friction = 100.4 N
So the friction force acting on the block is 100.4 N.
Finally, we can use Newton's second law (F = ma) to find the acceleration of the block:
F_net = F_applied - F_friction
F_net = 100 N - 100.4 N
F_net = -0.4 N (since the friction force is acting in the opposite direction to the applied force)
ma = -0.4 N (since the acceleration is in the opposite direction to the applied force)
a = -0.4 N / 10 kg
a = -0.04 m/s^2
So the block is decelerating at a rate of 0.04 m/s^2.
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A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer?
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A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer?.
A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer?.
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A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A 100 N force acts horizontally on a block of 10 kg placed on a horizontal rough surface of coefficient of friction μ = 0.5. If the acceleration due to gravity (g) is taken as 10 ms–2, the acceleration of the block (in ms–2) is [2002]a)2.5b)10c)5d)7.5Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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