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

Question

1447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-12T00:32:16

Approximately #0.14# joules.

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)#