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Given k1=1500Nm^-1, k2=500Nm^-1, m1=2kg,...

Given `k_1=1500Nm^-1`, `k_2=500Nm^-1`, `m_1=2kg`, `m_2=1kg`. Find:

a. potential energy stored in the spring in equilibrium, and
b. work done in slowly pulling down `m_2` by `8cm`.

Text Solution

Verified by Experts

The correct Answer is:
(a) `0.4` J, (b) 1 J
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