An 8.00 N weight is attached to the lower end of a spring which is fixed at its upper end. The
weight is initially held at rest at position X and the spring is unstretched. The weight is then
released and falls to position Y, which is 4.00 cm below X. The weight oscillates and then
eventually comes to rest at O, which is 2.00 cm below X.
Answers
Answered by
4
What should we have to find????
Answered by
13
Answer:B
Explanation:
At Y (position after initial drop), assume that the extension is 2x. The energy stored has the FORMULA= 1/2 x F (force in N) x e (extension in m only). Thus, the energy stored at Y is 1/2 x 8.0N x (4.0 x 10^ -3)m = 0.16J. Whereas at position O (after multiple oscillations), assume that the extension is 1x. Hence, the energy stored = 1/2 x 8.0 x (2.0x10^ -3)m = 0.08J. What we have to find is the Total Energy Lost from the system= 0.16J-0.08J = 0.08J which is the answer B.
A 0.04J B 0.08J C 0.16J D 0.32J
9702/11/O/N/11 QUESTION NUMBER 16.
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