(Assume Hooke\'s law applies.) It takes 50 J of work to stretch a spring 0.2 m f
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Question
(Assume Hooke's law applies.) It takes 50 J of work to stretch a spring 0.2 m from its equilibrium position. How much work is needed to stretch it an additional 0.5 m? Create a simple picture to illustrate the problem. Show your work; show all definite integrals that you evaluate. Round your final answer to 4 decimal positions. It takes 50 N of force to stretch a spring 0.2 m from its equilibrium position. How much work is needed to stretch it an additional 0.5 m? Create a simple picture to illustrate the problem. Show your work; show all definite integrals that you evaluate. Round your final answer to 4 decimal positions.Explanation / Answer
Ans.1
W = 1/2 kx²
k = 2 W/x²
k = (2*50J)/(0.2m)² = 2500 N/m
W = 1/2 kx² = 1/2 *2500N/ m*(0.5m)² = 312.5 J
Ans.2
Use Hooke's law F = - k x to determine the spring constant.
The spring constant k has the unit N/m, so k = 50N/0.2m = 250 N/m
Spring energy: E = 1/2 kx²
Work required to stretch 0.5m is
W = 1/2 * 250N/m*(0.5m)² = 31.25 J
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