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.