
When a mass $M$ is attached to the spring of force constant $k$, the spring stretches by $l$. If the mass oscillates with amplitude $l$, what will be the maximum potential energy stored in the spring?
A) $\dfrac{{kl}}{2}$
B) $2kl$
C) $\dfrac{1}{2}Mgl$
D) $Mgl$
Answer
164.4k+ views
Hint:
When a spring is stretched beyond its original length, a force must be given to it; this force relies on the spring constant and the length to which it has been stretched. Due to the length's expansion the potential energy will get stored inside the spring due to work done by this force.
Formula used:
The potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$
Complete step by step solution:
In order to know that a spring is stretched and extended by a certain amount of length when force is applied to it. $F = kx$ represents the force acting on the spring where $x$ is the spring's expanded extra length after being stretched, and $k$ is a constant.
According to the question, a mass of the spring$M$ is attached to the force which is constant $k$and the spring stretches by $l$. That means, $x = l$
Now, substitute the value of $x = l$in the above formula, then we have:
$F = kl$
Potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know, $Mg$force will act downward, so: $Mg = kl$
Therefore, the potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$
$\Rightarrow U = \dfrac{1}{2}kl \times l \\$
$\Rightarrow U = \dfrac{{Mgl}}{2} \\$
Thus, the correct option is:(C) $\dfrac{1}{2}Mgl$
Note:
It should be noted that stretching a spring results in the storage of potential energy. The work required to lengthen the spring is equivalent to potential energy. The force needed to extend a spring varies with distance, hence an integral is used in the work calculation.
When a spring is stretched beyond its original length, a force must be given to it; this force relies on the spring constant and the length to which it has been stretched. Due to the length's expansion the potential energy will get stored inside the spring due to work done by this force.
Formula used:
The potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$
Complete step by step solution:
In order to know that a spring is stretched and extended by a certain amount of length when force is applied to it. $F = kx$ represents the force acting on the spring where $x$ is the spring's expanded extra length after being stretched, and $k$ is a constant.
According to the question, a mass of the spring$M$ is attached to the force which is constant $k$and the spring stretches by $l$. That means, $x = l$
Now, substitute the value of $x = l$in the above formula, then we have:
$F = kl$
Potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know, $Mg$force will act downward, so: $Mg = kl$
Therefore, the potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$
$\Rightarrow U = \dfrac{1}{2}kl \times l \\$
$\Rightarrow U = \dfrac{{Mgl}}{2} \\$
Thus, the correct option is:(C) $\dfrac{1}{2}Mgl$
Note:
It should be noted that stretching a spring results in the storage of potential energy. The work required to lengthen the spring is equivalent to potential energy. The force needed to extend a spring varies with distance, hence an integral is used in the work calculation.
Recently Updated Pages
Environmental Chemistry Chapter for JEE Main Chemistry

Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
