Question

find the spring constant if a force of 1800 J of work to stretch a spring from it's natural length of 2M to a length of 5M?
Please quickly tommorow will my exam

Answer 1

can u plesase explain it more to understand me

Answer 2

I never get why it is negative, but it is the formula so

Answer 3

oh ok first you reply with positive now it's fine thanks very much...

Answer 4

It's not necessarily negative, but you usually set the reference axis so that the positive direction is from the spring towards the object it's pulling. The spring pulls the object, the force from the spring has therefore a direction in opposition with the positive direction of the axis, that's why it's negative.
The formulas used are correct but need to be a bit more rigurous.
I think you mistook for instance force and work, because you said the spring has a "force" of 1800 joules, while joule means work and not force. If you meant an actual force then the answer is correct... Otherwise:
Work being W
\[W=K/2 \times \left( x _{2}^{2}-x_{1}^{2} \right)\]
So
\[K=2W/( x _{2}^{2}-x_{1}^{2})\]
With xsub1 and xsub2 being the positions from which you lenghten the spring. However you lengthen it from its initial position by 3 meters so you have
\[K=2 \times 1800 / 3^{2}\]
So
\[K=400 N.m ^{-1}\]
With K=spring constant