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

1 Answer
Jun 12, 2018

Answer:

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)#