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

Answer: (D)

$25\,U$

Answer 2

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 $\frac{U'}{U}=25$
$\Rightarrow U'=25 U$