# If a spring has a constant of 4 (kg)/s^2, how much work will it take to extend the spring by 73 cm ?

Mar 24, 2018

1.0658 J

#### Explanation:

Spring constant, $k = 4 \frac{k g}{s} ^ 2$
Spring initial position, ${x}_{i} = 0 m$
Spring final position, ${x}_{f} = 0.73 m$

Work done by a spring can be found using the equation:
${W}_{s p} = \frac{1}{2} k \left({X}_{f}^{2} - {X}_{i}^{2}\right)$

${W}_{s p} = \frac{1}{2} \left(4\right) \left({\left(0.73\right)}^{2} - {\left(0\right)}^{2}\right)$
${W}_{s p} = 1.0658$ J (Don't forget about units)