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NEXT Chapter 10, Problem 38 GO Your answer is partially correct. Try again. A 0.45-kg metal sphere oscilates at the end of a vertical spring. As the spring stretches from 0.12 m to 0:23 m (relative to its unstrained length), the speed of the sphere decreases from 7.1 to 4.7 m/s. What is the spring constant of the spring? Number Units N/m the tolerance is +/-296 Click if you would like to Show Work for this question: Open Show Work

Answer 1

Answer :

Take gravitational potential energy when the spring is stretched by 0.23 m as 0
When the spring is stretched by 0.12 m, then it is at height h = 0.23 - 0.12 = 0.11 m above the point when it is stretched by 0.23 m

By conservation of energy
m * g * h1 + 1/2 * k * x1^2 + 1/2 * m * v1^2 = 0 + 1/2 * k * x2^2 + 1/2 * m * v2^2
0.45 * 9.8 * 0.11 + 1/2 * k * 0.12^2 + 1/2 * 0.45 * 7.1^2 = 1/2 * k * 0.23^2 + 1/2 * 0.45 * 4.7^2
0.4851 + 0.0072 k + 11.3423 = 0.02645 k + 4.970
0.0072 k + 11.82735 = 0.02645 k + 4.970
0.02645 k - 0.0072 k = 11.82735 - 4.970  
0.01925 k = 6.85735
k = 4.41452/0.012
k = 356.23 N/m
Ans: 356 N/m