If a spring has a constant of #8 (kg)/s^2#, how much work will it take to extend the spring by #40cm #?

1 Answer
Mar 18, 2018

Consider your system,

philschatz.com

Recall Hooke's Law,

#F_"s" = -kx#, and by extension

#F_"ext" = -F_"s" = kx# (we're looking to find this)

where #k# is the constant you describe and #x# is the displacement of spring.

Moreover, recall that work is represented by,

#W = F*d#

Now,

#=> W = (1/2kx)*x = 1/2kx^2#

(the reason this is halved is because we're averaging the force)

Hence,

#W_"ext" = 1/2 * (8"kg")/("s"^2) * (0.40"m")^2 approx 0.64"J"#

To be sure, the elastic potential energy is equal to the work done to stretch it that far (which shares the same equation as that derived).