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A spring with the spring constant k when compressed by 1 cm the potential energy stored is U if it is further compressed by 3 cm then change in its potential energy is (1) 3U (2) 9U (3) 8U (4) 15 U?
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A spring with the spring constant k when compressed by 1 cm the potent...
Explanation:

Given spring constant k and when compressed by 1 cm, potential energy stored is U.

Formula for potential energy stored in a spring is:

U = (1/2) kx^2

where x is the displacement from its equilibrium position.

When compressed by 1 cm, the displacement is x = -1 cm = -0.01 m.

So, U = (1/2) k (-0.01)^2 = (1/2) k (0.0001) = 0.00005 k


When further compressed by 3 cm,

The new displacement is x = -4 cm = -0.04 m.

The change in potential energy would be:


ΔU = U(final) - U(initial)

ΔU = (1/2) k(-0.04)^2 - (1/2) k(-0.01)^2

ΔU = (1/2) k (0.0016 - 0.0001)

ΔU = (1/2) k (0.0015)

ΔU = 0.00075 k


Answer:

So, the change in potential energy is 0.00075 k, which is not one of the given options. However, we can express it in terms of U:


ΔU = 0.00075 k = 0.00075 (1/U) U^2 = (0.00075/U) U^2

Since U = 0.00005 k, we have:

ΔU = (0.00075/U) U^2 = (0.00075/0.00005) U = 15 U


Therefore, the answer is option (4) 15U.
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A spring with the spring constant k when compressed by 1 cm the potential energy stored is U if it is further compressed by 3 cm then change in its potential energy is (1) 3U (2) 9U (3) 8U (4) 15 U?
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