Differculty working out what is doing work in a mass attached to a spring system

If you apply a force of 50N to the mass over the distance 1.67m, the mass won't be at rest after it's been lifted 1.67 meters. It will be moving rapidly upwards, with a bunch of extra kinetic energy and will overshoot the new equilibrium point by a fair amount.

Consider the situation before you apply the upwards force. If you apply even the most minuscule amount of force, just a tiny fraction of 50N, the mass will move up just a bit. The spring tension balanced the force of gravity; you added an additional upwards force so now the net force on the mass is upwards and it starts to accelerate upwards. The only reason it doesn't keep on moving upwards (as it would if you were using a counterweight instead of a spring) is that as it moves up the spring shortens so is exerting less upwards force and it reaches a new equilibrium. Increase the force a bit more, and you'll get it a bit further up... and so one until you're applying a force of 50N after lifting it 1.67 meters.

So it doesn't take 50N over the entire 1.67 distance to get the object to the new equilibrium point, and if you do apply 50N over that entire distance... You're throwing it upwards, past the equilibrium point.

 

yes. The force you'd be be applying at the start of the move would be 0N and and at the end it would be 50N.

To really do it right, you'd have to integrate [tex]W=\int{F(x)dx}[/tex] from 0 to 1.67, where F(x) is the force required to support the weight with the help of the spring after it's been lifted by an amount x.

But if you have a perfect spring, as this problem assumes, it turns out that the value of this integral is just [itex]\frac{XF(X)}{2}=\frac{Fd}{2}[/itex] where X is the total amount you've lifted the thing. And that's why you were off by a factor of two when you calculated W=Fd.