# If a spring has a constant of 6 (kg)/s^2, how much work will it take to extend the spring by 50 cm ?

Jan 28, 2016

The given potential energy of a spring can be found by the equation $P . E = \setminus \frac{1}{2} k {x}^{2}$

Given $k = 6 k g {s}^{-} 2$
Extension from previous point is $x = 0.5 m$.

Since we have not been provided with initial extension of the spring, we'll assume that the spring was at equilibrium initially with no potential energy. So the total potential energy of the spring will be the potential energy of spring due to this $0.5 m$ extension.

Now, $P . E = \setminus \frac{1}{2} k {x}^{2} = \setminus \frac{1}{2} \cdot 6 \cdot {0.5}^{2} = 3 \cdot 0.25 = 0.75 J$

So this will be the amount of work one has to do to stretch the spring by $0.5 m$