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Use g=10ms−2. A 2 kg mass is hung from an ideal spring with spring constant 40 N/m. The spring is then pulled down 20 cm from its equilibrium position. The work done in J due to the extension from equilibrium is closest to: |
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Answer» Work done = \(\int \vec F \cdot d\vec r\) Let y positive direction be pointing towards ground. Then: \(\vec F_1=-ky\hat j\) \(\vec F_2 = mg \hat j\) \(d\vec r = dy \hat j\) \(20cm = 0.2m\) Work is then \(\int_0^{0.2}mgdy - \int_0^{0.2}kydy\) = \(0.2m \cdot 10\frac{m}{s^2}\cdot 2kg - 40\frac{N}{m}\cdot(0.2m)^2\cdot 0.5 = 4J - 0.8J = 3.2J\) |
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