Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

16U

Step-by-Step Solution

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].
Establish Relationship: Since $k$ is constant for a given spring, the potential energy is directly proportional to the square of the extension ( $U \propto x^2$ ).
Calculate Ratio: Given $U_1 = U$ for $x_1 = 2\text{ cm}$ . We need to find $U_2$ for $x_2 = 8\text{ cm}$ . $\frac{U_2}{U_1} = \frac{\frac{1}{2}kx_2^2}{\frac{1}{2}kx_1^2} = \left(\frac{x_2}{x_1}\right)^2$ $\frac{U_2}{U} = \left(\frac{8}{2}\right)^2 = (4)^2 = 16$
Final Result: $U_2 = 16U$

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