If a spring has a constant of #7 (kg)/s^2#, how much work will it take to extend the spring by #41 cm #?

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Answer 1 · 2017-02-27T11:04:18

The work is #=0.59J#

The work done on a spring is

#W=1/2kx^2#

The elongation is #x=0.41m#

The spring constant is #=k=7kgs^-2=7Nm^-1#

The work is

#W=1/2*7*0.41*0.41#

#=0.59J#