A force of 800 N stretches a certain spring a distance of 0.200 m. (a) What is the potential energy of the spring when it is stretched 0.200 m? (b) What is its potential energy when it is compressed 5.00 cm?
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Solution 15E The force constant of a spring can be calculated from the equation k = F , where F is the x force applied on the spring and x is its elongation/compression. 1 2 The potential energy of a spring is given by P = kx 2 Now, given that F = 800 N x = 0.200 m Therefore, spring constant k = F x k = 800 N/0.200 m = 4000 N/m (a) otential energy of the spring when it is stretched by 0.200 m is, P = 1× 4000 N/m × (0.200) m 2 2 2 P = 80 J Therefore, the spring will have a potential energy of 80 J after stretching by 0.200 m. (b) Compression = 5.00 cm = 5.00/100 m = 0.05 m Potential energy after compression = 1 × 4000 N/m × (0.05) m = 5 J 2 Therefore, the spring will have a potential energy of 5 J after compressing by 5.00 cm.
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
The answer to “A force of 800 N stretches a certain spring a distance of 0.200 m. (a) What is the potential energy of the spring when it is stretched 0.200 m? (b) What is its potential energy when it is compressed 5.00 cm?” is broken down into a number of easy to follow steps, and 41 words. The full step-by-step solution to problem: 15E from chapter: 7 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: Energy, spring, potential, its, Force. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 15E from 7 chapter was answered, more than 293 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460.