If a spring has a constant of #2 (kg)/s^2#, how much work will it take to extend the spring by #56 cm #?

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Answer 1 · 2018-05-27T02:17:02

Around #0.31# joules.

Work on a spring is given by:

#W=1/2kx^2#

where:

#W# is the work done in joules

#k# is the spring constant

#x# is the extension, usually in meters

#:.x=56 \ "cm"=0.56 \ "m"#

So, the work done is:

#W=1/2*2 \ "kg/s"^2*(0.56 \ "m")^2#

#~~0.31 \ "kg m"^2"/s"^2#

#=0.31 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)#