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A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?
Most Upvoted Answer
A 0.5 kg block attached to an ideal spring with spring constant 80 New...
This is SHM.
k = 80 N/m m = 0.50 kgω = angular frequency of SHM of spring mass system=> ω² = k/m = 160=> ω = 4√10 rad/sec²Let the equation of SHM be: x = x₀ Sin (4√10 t) and v = x₀ * 4√10 Cos( 4√10 t)
Given that displacement 0.04 meters = x₀ Sin(4√10 t)and speed: 0.50 m/s = x₀ 4√10 * Cos (4√10 t)
Hence, 0.04² + (0.50/4√10)² = x₀² x₀ = 0.0562 meters or 5.62 cm
Thus the maximum speed of the block = x₀ * ω = 0.711 m/sec.....
k = 80 N/m m = 0.50 kgω = angular frequency of SHM of spring mass system=> ω² = k/m = 160=> ω = 4√10 rad/sec²Let the equation of SHM be: x = x₀ Sin (4√10 t) and v = x₀ * 4√10 Cos( 4√10 t)
Given that displacement 0.04 meters = x₀ Sin(4√10 t)and speed: 0.50 m/s = x₀ 4√10 * Cos (4√10 t)
Hence, 0.04² + (0.50/4√10)² = x₀² x₀ = 0.0562 meters or 5.62 cm
Thus the maximum speed of the block = x₀ * ω = 0.711 m/sec.....
Community Answer
A 0.5 kg block attached to an ideal spring with spring constant 80 New...
Introduction:
The problem states that a 0.5 kg block is attached to an ideal spring with a spring constant of 80 N/m. The block oscillates on a horizontal frictionless surface, and when the spring is compressed by 4 cm from its equilibrium position, the block has a speed of 0.5 m/s. We need to find the greatest speed of the block during its oscillation.
Key Information:
- Mass of the block (m) = 0.5 kg
- Spring constant (k) = 80 N/m
- Displacement of the spring from equilibrium (x) = 4 cm = 0.04 m
- Speed of the block (v) = 0.5 m/s
Analysis:
To find the greatest speed of the block, we need to consider the conservation of mechanical energy. In simple harmonic motion, the total mechanical energy remains constant.
The total mechanical energy of the system is given by the sum of the potential energy stored in the spring and the kinetic energy of the block:
E = PE + KE
1. Potential Energy (PE):
The potential energy stored in the spring is given by the formula:
PE = (1/2)kx²
Substituting the given values, we can calculate the potential energy:
PE = (1/2) * 80 * (0.04)²
PE = 0.064 J
2. Kinetic Energy (KE):
The kinetic energy of the block is given by the formula:
KE = (1/2)mv²
Substituting the given values, we can calculate the kinetic energy:
0.5 = (1/2) * 0.5 * v²
v² = 1
v = 1 m/s
3. Total Mechanical Energy:
Since the total mechanical energy remains constant, we can equate the potential energy and kinetic energy:
PE = KE
0.064 = (1/2) * 0.5 * v²
0.064 = 0.25 * v²
Solving for v, we get:
v² = 0.256
v = √0.256
v ≈ 0.506 m/s
Conclusion:
Therefore, the greatest speed of the block during its oscillation is approximately 0.506 m/s. The block reaches this maximum speed when the spring is at its equilibrium position.
The problem states that a 0.5 kg block is attached to an ideal spring with a spring constant of 80 N/m. The block oscillates on a horizontal frictionless surface, and when the spring is compressed by 4 cm from its equilibrium position, the block has a speed of 0.5 m/s. We need to find the greatest speed of the block during its oscillation.
Key Information:
- Mass of the block (m) = 0.5 kg
- Spring constant (k) = 80 N/m
- Displacement of the spring from equilibrium (x) = 4 cm = 0.04 m
- Speed of the block (v) = 0.5 m/s
Analysis:
To find the greatest speed of the block, we need to consider the conservation of mechanical energy. In simple harmonic motion, the total mechanical energy remains constant.
The total mechanical energy of the system is given by the sum of the potential energy stored in the spring and the kinetic energy of the block:
E = PE + KE
1. Potential Energy (PE):
The potential energy stored in the spring is given by the formula:
PE = (1/2)kx²
Substituting the given values, we can calculate the potential energy:
PE = (1/2) * 80 * (0.04)²
PE = 0.064 J
2. Kinetic Energy (KE):
The kinetic energy of the block is given by the formula:
KE = (1/2)mv²
Substituting the given values, we can calculate the kinetic energy:
0.5 = (1/2) * 0.5 * v²
v² = 1
v = 1 m/s
3. Total Mechanical Energy:
Since the total mechanical energy remains constant, we can equate the potential energy and kinetic energy:
PE = KE
0.064 = (1/2) * 0.5 * v²
0.064 = 0.25 * v²
Solving for v, we get:
v² = 0.256
v = √0.256
v ≈ 0.506 m/s
Conclusion:
Therefore, the greatest speed of the block during its oscillation is approximately 0.506 m/s. The block reaches this maximum speed when the spring is at its equilibrium position.
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A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?
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A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?.
A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 0.5 kg block attached to an ideal spring with spring constant 80 Newton per metre oscillates on a horizontal frictionless surface when the spring is 4 cm shorter than its equilibrium the speed of the block is point 0.5 M per second. the greatest speed of the block is?.
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