If a spring has a constant of #1 (kg)/s^2#, how much work will it take to extend the spring by #15 cm #?

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Answer 1 · 2018-05-20T01:53:44

Approximately #0.01# joules.

Work exerted on a spring is given by:

#W=1/2kx^2#

where:

#k# is the spring constant, usually in newtons per meter or kilograms per second squared.

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

Here: #x=15 \ "cm"=0.15 \ "m"#.

#:.W=1/2*1 \ "kg/s"^2*(0.15 \ "m")^2#

#=1/2 \ "kg/s"^2*0.0225 \ "m"^2#

#=0.01125 \ "kg m"^2"/s"^2#

#~~0.01 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)#