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

1 Answer
May 27, 2018

Answer:

Around #0.31# joules.

Explanation:

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)#