If a spring has a constant of #3 (kg)/s^2#, how much work will it take to extend the spring by #37 cm #?

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Answer 1 · 2018-05-16T13:30:47

Approximately #0.07# joules.

Work done on a spring is given by:

#W=1/2kx^2#

where:

#k# is the spring constant, usually in newtons per meter

#x# is the extension of the spring, usually in meters

So here: #x=37 \ "cm"=0.37 \ "m"#.

And so,

#W=1/2*3 \ "kg s"^-2*(0.37 \ "m")^2#

#=1/2*3 \ "kg s"^-2*0.1369 \ "m"^2#

#=1/2*3*0.1369 \ "J" \ (because 1 \ "J"-=1 \ "kg m"^2 \ "s"^-2)#

#=0.06845 \ "J"#

#~~0.07 \ "J"#