If a spring has a constant of #2 (kg)/s^2#, how much work will it take to extend the spring by #44 cm #?

1 Answer
Mar 20, 2018

Answer:

#"0.1936 Joule"#

Explanation:

Potential energy #("U")# gained by spring on stretching is

#"U" = 1/2\ "kx"^2#

#"U" = 1/2 × "2 kg/s"^2 × ("0.44 m")^2 = "0.1936 Joule"#

From work energy theorem, this equal to the amount of work done on the spring to stretch it by #"44 cm"#