Answer
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Hint: As the hey given force is constant, find out the work done in terms of force acting on the springs. As the force is constant and the only terms are work done on the spring and spring constants, we can derive a relation between the work done and the spring constant. it is given that work done on spring one is greater than the spring 2, therefore we can find the relation between the force constants.
Formula used:
$W=\dfrac{1}{2}k{{x}^{2}}$
Complete answer:
Let us find the forces acting on both the springs,
$F={{k}_{1}}{{x}_{1}}={{k}_{2}}{{x}_{2}}$
As the force is constant, the work done can be written in terms of force as,
$\begin{align}
& {{W}_{1}}=\dfrac{{{F}^{2}}}{2{{k}_{1}}} \\
& \Rightarrow {{W}_{2}}=\dfrac{{{F}^{2}}}{2{{k}_{2}}} \\
\end{align}$
Now, it is given that the work done by spring one is greater than the spring two.
$\begin{align}
& {{W}_{1}}>{{W}_{2}} \\
& {{k}_{1}}<{{k}_{2}} \\
\end{align}$
Therefore, statement one is false, and it tells us that work done by spring two is greater than spring one. Statement two is true.
Additional Information:
Elastic potential energy is the potential energy stored because of deformation of an elastic object, such as stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant as well as the distance stretched. Hey according to Hooke’s law, the force required to stretch the spring will be directly proportional to the amount of stretch. Springs are very important because they serve as simple models for lots of complicated physical systems. Objects which behave like springs behave in a manner which is described as simple harmonic motion simple harmonic motion plays a major role in physics. The defining character of spring is that it resists displacement from its rest position with a force which increases linearly.
Note:
In the above question, the force acting on the springs is constant. Therefore, the work done is formulated in terms of spring constant and the distance travelled by the springs. The work done in spring is inversely proportional to the spring constant. The more the spring constant the less the work is done when this force acting is constant.
Formula used:
$W=\dfrac{1}{2}k{{x}^{2}}$
Complete answer:
Let us find the forces acting on both the springs,
$F={{k}_{1}}{{x}_{1}}={{k}_{2}}{{x}_{2}}$
As the force is constant, the work done can be written in terms of force as,
$\begin{align}
& {{W}_{1}}=\dfrac{{{F}^{2}}}{2{{k}_{1}}} \\
& \Rightarrow {{W}_{2}}=\dfrac{{{F}^{2}}}{2{{k}_{2}}} \\
\end{align}$
Now, it is given that the work done by spring one is greater than the spring two.
$\begin{align}
& {{W}_{1}}>{{W}_{2}} \\
& {{k}_{1}}<{{k}_{2}} \\
\end{align}$
Therefore, statement one is false, and it tells us that work done by spring two is greater than spring one. Statement two is true.
Additional Information:
Elastic potential energy is the potential energy stored because of deformation of an elastic object, such as stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant as well as the distance stretched. Hey according to Hooke’s law, the force required to stretch the spring will be directly proportional to the amount of stretch. Springs are very important because they serve as simple models for lots of complicated physical systems. Objects which behave like springs behave in a manner which is described as simple harmonic motion simple harmonic motion plays a major role in physics. The defining character of spring is that it resists displacement from its rest position with a force which increases linearly.
Note:
In the above question, the force acting on the springs is constant. Therefore, the work done is formulated in terms of spring constant and the distance travelled by the springs. The work done in spring is inversely proportional to the spring constant. The more the spring constant the less the work is done when this force acting is constant.
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