The potential energy of a long spring when stretched by $2 ext{ cm}$ is $U$. If the spring is stretched by $8

Question

The potential energy of a long spring when stretched by $2	ext{ cm}$ is $U$. If the spring is stretched by $8	ext{ cm}$, the potential energy stored in it will be:

Accepted Answer

1. **Identify the Formula:** The potential energy ($U$) stored in a spring of spring constant $k$ stretched by a distance $x$ is given by $U = \frac{1}{2}kx^2$ [Class 11 Physics, Ch 5, Sec 5.9, Eq 5.15].
2. **Establish Relationship:** Since $k$ is constant for a given spring, the potential energy is directly proportional to the square of the extension ($U \propto x^2$).
3. **Calculate Ratio:**
   Given $U_1 = U$ for $x_1 = 2	ext{ cm}$.
   We need to find $U_2$ for $x_2 = 8	ext{ cm}$.
   $$ \frac{U_2}{U_1} = \frac{\frac{1}{2}kx_2^2}{\frac{1}{2}kx_1^2} = \left(\frac{x_2}{x_1}ight)^2 $$
   $$ \frac{U_2}{U} = \left(\frac{8}{2}ight)^2 = (4)^2 = 16 $$
4. **Final Result:**
   $$ U_2 = 16U $$

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