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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.?
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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 k...
**Solution:**
To solve this problem, we will analyze the forces acting on the system and apply Newton's laws of motion.
**1. Analysis of the forces:**
- The 3 kg block is resting on a horizontal surface, so the normal force acting on it is equal to its weight, which is given by N = mg, where m = 3 kg and g = 9.8 m/s^2 (acceleration due to gravity).
- The 2 kg block is in equilibrium, which means that the force pulling it downwards (its weight) is balanced by the force pulling it upwards (the tension in the spring). Therefore, the tension in the spring is also equal to mg.
**2. Applying Newton's laws of motion:**
- For the 3 kg block:
- The only horizontal force acting on it is the tension in the spring. Therefore, we have T = ma, where T is the tension, m is the mass, and a is the acceleration.
- Since the block is at rest, the acceleration is zero. Therefore, the tension in the spring is also zero.
- For the 2 kg block:
- The vertical forces acting on it are its weight (mg) and the tension in the spring.
- The net force in the vertical direction is given by F_net = mg - T.
- Since the block is in equilibrium, the net force is zero. Therefore, we have mg - T = 0.
- Rearranging the equation, we get T = mg.
**3. Calculating the compression of the spring:**
- The compression of the spring is given by x = x_0 + x_1, where x_0 is the additional compression required and x_1 is the initial compression of 1 cm.
- From the previous analysis, we know that the tension in the spring is mg.
- The force exerted by the spring is given by Hooke's law, F = kx, where F is the force, k is the spring constant, and x is the compression.
- Equating the force exerted by the spring to the tension, we have kx = mg.
- Solving for x, we get x = mg/k.
- Substituting the values, we have x = (2 kg * 9.8 m/s^2) / k.
Therefore, the additional compression required, x_0, is equal to (2 kg * 9.8 m/s^2) / k.
To solve this problem, we will analyze the forces acting on the system and apply Newton's laws of motion.
**1. Analysis of the forces:**
- The 3 kg block is resting on a horizontal surface, so the normal force acting on it is equal to its weight, which is given by N = mg, where m = 3 kg and g = 9.8 m/s^2 (acceleration due to gravity).
- The 2 kg block is in equilibrium, which means that the force pulling it downwards (its weight) is balanced by the force pulling it upwards (the tension in the spring). Therefore, the tension in the spring is also equal to mg.
**2. Applying Newton's laws of motion:**
- For the 3 kg block:
- The only horizontal force acting on it is the tension in the spring. Therefore, we have T = ma, where T is the tension, m is the mass, and a is the acceleration.
- Since the block is at rest, the acceleration is zero. Therefore, the tension in the spring is also zero.
- For the 2 kg block:
- The vertical forces acting on it are its weight (mg) and the tension in the spring.
- The net force in the vertical direction is given by F_net = mg - T.
- Since the block is in equilibrium, the net force is zero. Therefore, we have mg - T = 0.
- Rearranging the equation, we get T = mg.
**3. Calculating the compression of the spring:**
- The compression of the spring is given by x = x_0 + x_1, where x_0 is the additional compression required and x_1 is the initial compression of 1 cm.
- From the previous analysis, we know that the tension in the spring is mg.
- The force exerted by the spring is given by Hooke's law, F = kx, where F is the force, k is the spring constant, and x is the compression.
- Equating the force exerted by the spring to the tension, we have kx = mg.
- Solving for x, we get x = mg/k.
- Substituting the values, we have x = (2 kg * 9.8 m/s^2) / k.
Therefore, the additional compression required, x_0, is equal to (2 kg * 9.8 m/s^2) / k.
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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 k...
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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.?
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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.?.
The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.?.
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The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.?, a detailed solution for The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? has been provided alongside types of The ends of spring are attached to blocks of mass 3kg and 2kg. The 3 kg block rests on a horizontal surface and the 2 kg block which is vertically above it is equilibrium producing a compression of 1 cm of the spring. The 2 kg mass must be compressed further by an amount x0, so that when it is released,the 3 kg block may be lifted off the ground. Find x0.? theory, EduRev gives you an
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