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A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s?
? Related: Newton's Laws of Motion
Most Upvoted Answer
A 60 kg painter on a 15 kg platform. A rope attached to the platform a...
**Solution:**
Given:
- Mass of the painter, m₁ = 60 kg
- Mass of the platform, m₂ = 15 kg
- Force exerted by the painter to start, F = 400 N
(i) **Acceleration of the platform and painter:**
To find the acceleration, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
1. Find the total mass of the system:
- Total mass, M = m₁ + m₂ = 60 kg + 15 kg = 75 kg
2. Find the net force acting on the system:
- Net force, F_net = F - Weight of the system
- Weight of the system = (m₁ + m₂) * g, where g is the acceleration due to gravity (9.8 m/s²)
- Weight of the system = 75 kg * 9.8 m/s² = 735 N
- F_net = 400 N - 735 N = -335 N (negative sign indicates downward force)
3. Calculate the acceleration:
- F_net = M * a
- -335 N = 75 kg * a
- a = -4.47 m/s² (negative sign indicates downward acceleration)
Therefore, the acceleration of the platform and painter is -4.47 m/s², directed downward.
(ii) **Force required to attain an upward speed of 1 m/s in 1 s:**
To find the force required, we need to apply Newton's second law of motion again, using the given time interval and desired velocity.
1. Calculate the change in velocity:
- Change in velocity, Δv = Final velocity - Initial velocity = 1 m/s - 0 m/s = 1 m/s
2. Find the acceleration:
- Acceleration, a = Δv / t = 1 m/s / 1 s = 1 m/s²
3. Find the net force required:
- F_net = M * a = 75 kg * 1 m/s² = 75 N
Therefore, the force required to attain an upward speed of 1 m/s in 1 s is 75 N.
(iii) **Force required to maintain a constant speed of 1 m/s:**
Once the desired speed of 1 m/s is achieved, the force required to maintain the constant speed is determined by the equilibrium condition. According to Newton's first law of motion, an object will remain at rest or move with constant velocity unless acted upon by a net external force.
In this case, since the speed is constant, the net force acting on the system must be zero.
1. Calculate the net force required to maintain constant speed:
- F_net = 0 N
2. Find the force exerted by the painter:
- F = Weight of the system
- Weight of the system = (m₁ + m₂) * g = 75 kg * 9.8 m/s² = 735 N
Therefore, the force that should be applied by the painter to maintain a constant speed of 1 m/s is 735 N.
Given:
- Mass of the painter, m₁ = 60 kg
- Mass of the platform, m₂ = 15 kg
- Force exerted by the painter to start, F = 400 N
(i) **Acceleration of the platform and painter:**
To find the acceleration, we need to apply Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.
1. Find the total mass of the system:
- Total mass, M = m₁ + m₂ = 60 kg + 15 kg = 75 kg
2. Find the net force acting on the system:
- Net force, F_net = F - Weight of the system
- Weight of the system = (m₁ + m₂) * g, where g is the acceleration due to gravity (9.8 m/s²)
- Weight of the system = 75 kg * 9.8 m/s² = 735 N
- F_net = 400 N - 735 N = -335 N (negative sign indicates downward force)
3. Calculate the acceleration:
- F_net = M * a
- -335 N = 75 kg * a
- a = -4.47 m/s² (negative sign indicates downward acceleration)
Therefore, the acceleration of the platform and painter is -4.47 m/s², directed downward.
(ii) **Force required to attain an upward speed of 1 m/s in 1 s:**
To find the force required, we need to apply Newton's second law of motion again, using the given time interval and desired velocity.
1. Calculate the change in velocity:
- Change in velocity, Δv = Final velocity - Initial velocity = 1 m/s - 0 m/s = 1 m/s
2. Find the acceleration:
- Acceleration, a = Δv / t = 1 m/s / 1 s = 1 m/s²
3. Find the net force required:
- F_net = M * a = 75 kg * 1 m/s² = 75 N
Therefore, the force required to attain an upward speed of 1 m/s in 1 s is 75 N.
(iii) **Force required to maintain a constant speed of 1 m/s:**
Once the desired speed of 1 m/s is achieved, the force required to maintain the constant speed is determined by the equilibrium condition. According to Newton's first law of motion, an object will remain at rest or move with constant velocity unless acted upon by a net external force.
In this case, since the speed is constant, the net force acting on the system must be zero.
1. Calculate the net force required to maintain constant speed:
- F_net = 0 N
2. Find the force exerted by the painter:
- F = Weight of the system
- Weight of the system = (m₁ + m₂) * g = 75 kg * 9.8 m/s² = 735 N
Therefore, the force that should be applied by the painter to maintain a constant speed of 1 m/s is 735 N.
Community Answer
A 60 kg painter on a 15 kg platform. A rope attached to the platform a...
1) 2T-(M+m)g=(M+m)a M=60kg m =15kg T=400 N g=10
substituting value
a= 0.67m/s2
2) F=1/2 (M+m)(g+a)
substituting values
F=412.5N
3)
2F-(M+m)g=0
putting value
F=375 N
hope it's clear
substituting value
a= 0.67m/s2
2) F=1/2 (M+m)(g+a)
substituting values
F=412.5N
3)
2F-(M+m)g=0
putting value
F=375 N
hope it's clear
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A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion?
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A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion?.
A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion?.
Solutions for A 60 kg painter on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself along with the platform. (i) To get started, he pulls the rope down with a force of 400 N. Find the acceleration of the platform as well as that of the painter.(ii) What force must he exert on the rope so as to attain an upward speed of 1 m/s in 1 s ?(iii) What force should apply now to maintain the constant speed of 1 m/s? Related: Newton's Laws of Motion? in English & in Hindi are available as part of our courses for Class 11.
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