## Question

###### If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #26 cm #?

If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #26 cm #?

## Answers

Approximately

#0.14# joules.## Explanation:

Work done on a spring is given by the equation:

#W=1/2kx^2# where:

#k# is the spring constant

#x# is the extension in meters

#:.x=26 \ "cm"=0.26 \ "m"# So the work done is:

#W=1/2*4 \ "kg/s"^2*(0.26 \ "m")^2#

#=2 \ "kg/s"^2*0.0676 \ "m"^2#

#~~0.14 \ "kg m"^2"/s"^2#

#=0.14 \ "J" \ (because 1 \ "J"=1 \ "kg m"^2"/s"^2)#

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