Physics Review Question Unit: WORK AND ENERGY Year: 1995 Question#: 17 Question: When a spring is stretched 0.200 meter from its equilibrium position, it possesses a potential energy of 10.0 joules. What is the spring constant for this spring? (1 ) 100. N/m (2 ) 125 N/m (3 ) 250. N/m (4 ) 500. N/m What is this question really asking? Explanation by: JD W. [ Return to question menu ] Answer 1 100. N/m INCORRECT: The equation for this spring constant problem is PE=1/2kx2. If you use this equation with the given information, the answer is not 100 N/m. return to top Answer 2 125 N/m INCORRECT INCORRECT: The equation for this spring constant problem is PE=1/2kx2. return to top Answer 3 250. N/m INCORRECT The equation for this spring constant problem is PE=1/2kx2. return to top Answer 4 500. N/m CORRECT Answer 4 is correct. the equation is PE=1/2kx2. SO: GIVEN: PE=10J x=0.2m PE=1/2kx2 10J=1/2(k)(0.2m)2 10J=0.02m2k k=500N/m return to top What's this question really asking? This question is really asking what the spring constant is. The spring constant is the ratio of the force required to stretch or compress a particular spring to the distance of stretch or compression. If you know the spring constant you can then apply the formula PE=1/2kx2. return to top |
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