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A block of mass 10 kg is moving in x-direction with a constant velocity of 10 m−s⁻1. It is subjected to a retarding force F = 0.1x J-m⁻1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will be
- a)475 J
- b)450 J
- c)275 J
- d)250 J
Correct answer is option 'A'. Can you explain this answer?
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A block of mass 10 kg is moving in x-direction with a constant velocit...
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A block of mass 10 kg is moving in x-direction with a constant velocit...
/s. If a force of 30 N is applied in the opposite direction of motion, what will be the new velocity of the block?
We can use the formula:
F = ma
where F is the force, m is the mass, and a is the acceleration.
Since the block is already moving with a constant velocity, we know that there is no net force acting on it. This means that the force of friction is equal and opposite to the applied force. So, we can write:
30 N - F_friction = 0
where F_friction is the force of friction.
Solving for F_friction, we get:
F_friction = 30 N
Now, we can use the formula:
F_friction = μ * N
where μ is the coefficient of friction and N is the normal force.
Since the block is moving horizontally, the normal force is equal to the weight of the block, which is:
N = m * g
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
So, we can write:
F_friction = μ * m * g
30 N = μ * 10 kg * 9.8 m/s^2
Solving for μ, we get:
μ = 0.306
Now, we can use the formula:
F = ma
where F is the net force, m is the mass, and a is the acceleration.
The net force is:
F = 30 N - F_friction
F = 30 N - μ * m * g
F = 30 N - 3 N
F = 27 N
The acceleration is:
a = F / m
a = 27 N / 10 kg
a = 2.7 m/s^2
The new velocity can be found using the formula:
v = v0 + at
where v0 is the initial velocity (10 m/s), a is the acceleration (-2.7 m/s^2), and t is the time taken.
Since the block is moving with a constant velocity, we know that the time taken for the force to change the velocity is zero. Therefore, the new velocity will be:
v = 10 m/s + 0
v = 10 m/s
So, the new velocity of the block is still 10 m/s, but the force of friction has increased to balance the applied force.
We can use the formula:
F = ma
where F is the force, m is the mass, and a is the acceleration.
Since the block is already moving with a constant velocity, we know that there is no net force acting on it. This means that the force of friction is equal and opposite to the applied force. So, we can write:
30 N - F_friction = 0
where F_friction is the force of friction.
Solving for F_friction, we get:
F_friction = 30 N
Now, we can use the formula:
F_friction = μ * N
where μ is the coefficient of friction and N is the normal force.
Since the block is moving horizontally, the normal force is equal to the weight of the block, which is:
N = m * g
where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.
So, we can write:
F_friction = μ * m * g
30 N = μ * 10 kg * 9.8 m/s^2
Solving for μ, we get:
μ = 0.306
Now, we can use the formula:
F = ma
where F is the net force, m is the mass, and a is the acceleration.
The net force is:
F = 30 N - F_friction
F = 30 N - μ * m * g
F = 30 N - 3 N
F = 27 N
The acceleration is:
a = F / m
a = 27 N / 10 kg
a = 2.7 m/s^2
The new velocity can be found using the formula:
v = v0 + at
where v0 is the initial velocity (10 m/s), a is the acceleration (-2.7 m/s^2), and t is the time taken.
Since the block is moving with a constant velocity, we know that the time taken for the force to change the velocity is zero. Therefore, the new velocity will be:
v = 10 m/s + 0
v = 10 m/s
So, the new velocity of the block is still 10 m/s, but the force of friction has increased to balance the applied force.
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A block of mass 10 kg is moving in x-direction with a constant velocity of 10 m−s⁻1. It is subjected to a retarding force F = 0.1x J-m⁻1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will bea)475 Jb)450 Jc)275 Jd)250 JCorrect answer is option 'A'. Can you explain this answer?
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A block of mass 10 kg is moving in x-direction with a constant velocity of 10 m−s⁻1. It is subjected to a retarding force F = 0.1x J-m⁻1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will bea)475 Jb)450 Jc)275 Jd)250 JCorrect answer is option 'A'. 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 block of mass 10 kg is moving in x-direction with a constant velocity of 10 m−s⁻1. It is subjected to a retarding force F = 0.1x J-m⁻1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will bea)475 Jb)450 Jc)275 Jd)250 JCorrect answer is option 'A'. 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 block of mass 10 kg is moving in x-direction with a constant velocity of 10 m−s⁻1. It is subjected to a retarding force F = 0.1x J-m⁻1 during its travel from x = 20 m to x = 30 m. Its final kinetic energy will bea)475 Jb)450 Jc)275 Jd)250 JCorrect answer is option 'A'. Can you explain this answer?.
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