
Two springs of spring constant $1500N/m$and 3000N /m respectively are stretched with the same force. They will have the potential energies in the ratio of (A) $1:9$(B) $1:4$(C) $2:1$(D) $4:1$
Answer
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Hint: The resistive force developed in the spring, when its length is changed it is called a spring force. Potential energy of a spring is the energy associated with the state of compression or expansion of an elastic spring. Spring force is usually given by $F{\text{ }} = - kx$. Here $k$ is the spring constant and $x$ is the displacement in the spring.
Complete step by step solution
Note: The energy possessed by a body or system by virtue of its position or configuration is known as the potential energy. For example, a block attached to a compressed or elongated spring possesses some energy called elastic potential energy. This block has a capacity to do work.
Complete step by step solution
When work is done in pulling the string, it gets stored as potential energy (U).
$\Rightarrow U = \dfrac{1}{2}\mathop {kx}\nolimits^2 … (1)$
The restoring force developed in spring is directly proportional to elongation or compression $x$
$\Rightarrow F{\text{ }} = {\text{ }}kx{\text{ }} … (2)$
Now, $x = \dfrac{F}{k}$
Putting this value in equation (1)
$\Rightarrow U = \dfrac{1}{2}\mathop {k\left( {\dfrac{F}{k}} \right)}\nolimits^2 $
Now, $\dfrac{{\mathop U\nolimits_1 }}{{\mathop U\nolimits_2 }} = \dfrac{{\mathop k\nolimits_2 }}{{\mathop k\nolimits_1 }}$
Now putting values of spring constants-
$\mathop K\nolimits_1 {\text{ }} = {\text{ }}1500{\text{ }}N/m$
$\mathop K\nolimits_2 {\text{ }} = {\text{ 30}}00{\text{ }}N/m$
$ \Rightarrow \dfrac{{3000}}{{1500}} = \dfrac{2}{1} = 2:1$
Therefore, Option (C) is correct.
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