A spring with k = 54 N/cm is initially stretched 1 cm from its equilibrium length.
(a) How much more energy is needed to further stretch the spring to 2 cm beyond its equilibrium length?
(b) From this new position, how much energy is needed to compress the spring to 2 cm shorter than its equilibrium position?
Physics - Elena, Saturday, June 2, 2012 at 12:19pm
PE2 = k•x2²/2
W = ΔPE =PE2 –PE1 =k(x2² - x1²)/2 = ….
where k =54 N/m, x1 =0.01 m,
x2 =0.01+0.02 = 0.03 m.
this compression will be done by the work of elastic force of the spring. When we release the spring which is stretched by 0.03 m, the released energy is k•(0.003)²/2 which is large than k•(0.02)²/2.
Physics - Kelly, Saturday, June 2, 2012 at 8:28pm
I'm still a little confused. When I use the equation:
W = k(x2^2 - x1^2)/2, I get 0.0216 J. This doesn't seem correct.