If a spring has a constant of #2 (kg)/s^2#, how much work will it take to extend the spring by #87 cm #?

1 Answer
Mar 14, 2018

Answer:

#"0.76 Joule"#

Explanation:

Energy stored in a spring is given by

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

Energy stored in spring when it’s streched by #"87 cm"# is

#U = 1/2\ "kx"^2 = 1/cancel(2)\ cancel(2)\ "kg/s"^2 × ("0.87 m")^2 = "0.76 Joule"#

Work = #U_f - U_i = "0.76 J - 0 J = 0.76 J"#