If a spring has a constant of #4 (kg)/s^2#, how much work will it take to extend the spring by #65 cm #?

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Answer 1 · 2018-04-11T13:57:38

The work done is #0.845# joules.

We use the work equation for a spring, which states that,

#W=1/2kx^2#

#k# is the spring constant in #"N/m"#

#x# is the extension of the spring in meters

Note that:

#1 \ "N/m"=("kg m""/s"^2)/("m")=1 \ "kg/s"^2#

#:.4 \ "kg/s"^2=4 \ "N/m"#

So, we have:

#65 \ "cm"=0.65 \ "m"#

And so, the work done is:

#W=1/2*4 \ "N/m"*(0.65 \ "m")^2#

#=2 \ "N/m"*0.4225 \ "m"^2#

#=0.845 \ "N m"#

#=0.845 \ "J"#