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Physics Question 6 – NEET-UG 2023

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

The potential energy stored in a spring is related to its extension or compression.

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Ninja StrategySquare Proportionality

Recognize that elastic potential energy is proportional to the square of the extension. If extension increases by a factor of 'n', energy increases by 'n^2'.

Step 1: Recall the formula for elastic potential energy✦ Active

The potential energy (U) stored in a spring is directly proportional to the square of its extension (x) or compression. The formula is:

U=12kx2

where k is the spring constant.

Step 2: Set up the initial condition○ Expand

Given that when the spring is stretched by x1=2 cm, the potential energy is U1=U. Using the formula:

U=12k(2 cm)2=12k(4 cm2)(1)
Step 3: Calculate the potential energy for the new extension○ Expand

When the spring is stretched by x2=8 cm, let the new potential energy be U2. Using the formula again:

U2=12k(8 cm)2=12k(64 cm2)(2)

Now, divide equation (2) by equation (1):

U2U=12k(64 cm2)12k(4 cm2)=644=16

Therefore, U2=16U.

💡 Teacher's Secret Hint

Notice that the units of extension (cm) cancel out in the ratio, so conversion to meters is not strictly necessary for this problem, but it's good practice for other calculations.

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