In this physics problem, we explore the spring-mass system where an 8.00 kg stone compresses a spring by 10.0 cm.
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The problem is broken down into four parts
(a) What is the spring constant?
(b) If the stone is pushed down further by 30.0 cm and then released, what is the elastic potential energy of the compressed spring just before release?
(c) What is the change in the gravitational potential energy of the stone as it moves from the release point to its maximum height?
(d) What is the maximum height, measured from the release point?
Solution Steps:
(a) Finding the Spring Constant:
Understanding the spring constant is crucial as it defines how stiff the spring is. By analyzing the spring-mass system in equilibrium, where the spring force balances the gravitational force on the stone, we use Newton's second law (F_net = ma) to find:
F_spring - mg = 0
-k(-0.100 m) - (8.00 kg)(9.8 m/s²) = 0
Solving this, we determine the spring constant k to be 784 N/m.
(b) Calculating the Elastic Potential Energy:
The stone is pushed further, compressing the spring by an additional 30.0 cm, leading to a total compression of 40.0 cm (0.400 m). The elastic potential energy stored in the spring at the moment of release is given by:
U = (1/2) k y₁²
U = (1/2) (784 N/m) (0.400 m)²
This calculation shows that the elastic potential energy is 62.7 J. This energy is crucial as it transforms into kinetic energy and subsequently into gravitational potential energy.
(c) Change in Gravitational Potential Energy:
As the stone moves from the release point to its maximum height, the gravitational potential energy changes. Using the principle of energy conservation between the release point (y₁) and maximum height (y₂), we write:
K₁ + U₁ = K₂ + U₂
At y₁, kinetic energy (KE) is zero, and potential energy (PE) is (1/2) k y₁².
At y₂, kinetic energy is again zero, and potential energy is mgh, where h = y₂ - y₁.
The change in gravitational potential energy equals the elastic potential energy at the release point, which is 62.7 J.
(d) Determining the Maximum Height:
The maximum height h from the release point can be calculated as:
h = (k y₁²) / (2mg)
h = 0.800 m (or 80.0 cm).
This represents the highest point the stone reaches due to the conversion of elastic potential energy into gravitational potential energy.
Key Takeaways:
Spring Constant: Understanding this fundamental property helps in analyzing the relationship between the force exerted by the spring and its displacement.
Elastic Potential Energy: This energy stored in the spring due to compression is vital in predicting the motion of objects in a spring-mass system.
Energy Conservation: The total mechanical energy (kinetic + potential) is conserved in the system, a key concept in understanding the interplay between different forms of energy in physics.
Who is this for: Class 11 physics student, AP Physics students, students appearing for competitive exams like IIT JEE and NEET. If you are looking for Class 11 physics notes or PDF downloads on the topic of work, energy and power, you should visit our website. You can find quiz and pdf summary of the topic.
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