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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