Vertically driven mass against spring

[rob78]

Please help!
Im not sure which is the correct theory to use for this problem.
A mass 1kg is driven vertically upwards against a spring. The spring has no initial compression and a spring rate of 400N/m. The mass will reach a height of 0.180m.
What is the input energy if ideal conditions are assumed here.
I have been given the formula E=mgS+kS²,
but do not understand why simply adding :
E(gravity)=mgS and E(spring)=½kS² do not give me the same value.
Further to this, what would be the effect of adding preload to the spring?

 

[Doc Al]

I have been given the formula E=mgS+kS²,

Looks like a typo to me.

but do not understand why simply adding :
E(gravity)=mgS and E(spring)=½kS² do not give me the same value.

I suspect that's what they meant to do.

 

[rob78]

Thanks,
The derivation for E=mgS+kS² went like this:
V²=U²+2aS, V=0
therefore,
U²=-2aS, where a=g(9.81)
KE=½mU²
Substituting gives:
KE=½m*(2gS) = mgS

adding in the spring, Fspring=kS
the deceleration will now be (g+[kS/m])
which will give:
KE=[g+(kS/m)]*mS = gms+kS²

working this through, I can't see any errors but maybe I am missing something?

 

[Doc Al]

The derivation for E=mgS+kS² went like this:
V²=U²+2aS, V=0
therefore,
U²=-2aS, where a=g(9.81)
KE=½mU²
Substituting gives:
KE=½m*(2gS) = mgS

Here only gravity acts. The acceleration is constant.

adding in the spring, Fspring=kS
the deceleration will now be (g+[kS/m])
which will give:
KE=[g+(kS/m)]*mS = gms+kS²

The error here is that the acceleration is no longer constant. Hint: What's the average acceleration as the spring compresses from x = 0 to x = S?

 

[rob78]

OK,
The average acceleration will be ½kS, meaning that this now gives the work done to compress the spring from rest for the second half, and gravitational energy for the first half.
Excellent, that's what I had expected but had missed the point of it being average. This now makes sense, MANY THANKS.