Question

A long spring is stretched by $ 2\,cm $ and its potential energy is $ U $ . If the spring is stretched by $ 10\,cm $ ; its potential energy will be

• A. $ U/5 $
• B. $ U/25 $
• C. $ 5\,U $
• D. $ 25\,U $

Answer 1

The Correct Option is D

The potential energy of a stretched spring is
$U=\frac{1}{2} k x^{2}$
Here, $k=$ spring constant,
$x=$ elongation in spring.
But given that, the elongation is $2 cm .$
So, $U =\frac{1}{2} k(2)^{2}$
$\Rightarrow U =\frac{1}{2} k \times 4$...(i)
If elongation is $10 cm$ then potential energy
$U'=\frac{1}{2} k(10)^{2}$
$U'=\frac{1}{2} k \times 100$...(ii)
On dividing E (ii) by E (i), we have
$\frac{U'}{U}=\frac{\frac{1}{2} k \times 100}{\frac{1}{2} k \times 4}$
or $\frac{U'}{U}=25$
$\Rightarrow U'=25 U$

Answer 2

So, here in the First case: potential energy of the spring is, 

\(U=\frac{kx^{2}}{2}\), where k is the spring constant of N/m and x is the displacement from the equilibrium position in metres. x = 2 cm = 0.02 m Therefore potential energy, \(U=\frac{k\times0.02^{2}}{2}\)

Therefore, k=5000U 

So now, for second case: x= 10cm = 0.1m 

\(U^{'}=\frac{kx'^{2}}{2}\) 

⇒\(U'=\frac{k\times0.1^{2}}{2}\) 

⇒ \(U'=\frac{5000U\times0.1^{2}}{2}\) 

Therefore, k = 25U 

So, the correct option is D.