# If a spring has a constant of 1 (kg)/s^2, how much work will it take to compress the spring 5 cm ?

Here work done = PE of compressed spring =$\frac{1}{2} k {x}^{2}$
Given $k = 1 \frac{k g}{s} ^ 2$
compression $x = 5 c m = 0.05 m$
work done = PE of compressed spring =$\frac{1}{2} k {x}^{2} = \frac{1}{2} \times 1 \times {\left(0.05\right)}^{2} = 0.0125 J$