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?

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.