How Much Work to Stretch a Spring 4 cm from Equilibrium?

[omega_minus]

From the sound of your question it sounds like you're still working the problem from the floor up. It's not a gravitational potential energy problem, as others have said. Work the problem from the ceiling down, from where the spring is hung. Those are the measurements that matter. The height above the floor is not important, but as Fewmet has said g will play a role in the calculation.

 

[flyingpig]

I see it now, I overlooked it.

\sum W = \int_{x_0 = 3.00cm}^{x = 4.00cm} -kx\cdot dx - mg(-1.00cm)

Right? I have negative change in distance because I am climbing "down hill"

 

[ideasrule]

If you compute that answer, you'll find that it's negative. That's because you calculated the work exerted by the spring on the external agent, not the work exerted by the external agent on the spring.

Here's a useful way to approach this problem. We're applying a force to pull the spring down, and that force is kx. Gravity provides some help, so we only really need to apply kx-mg. The integral of kx-mg from x=3cm to x=4cm is the work that the external agent needs to do.

 

[flyingpig]

Well we don't even know what k is

I calculated and I got

-3.5 * 10^-4 * k + 50 * 1cm= Work

 

[ideasrule]

The question tells you that when a 5kg mass is hung on the spring, it stretches by 3cm. A 5kg mass applies a force of (5 kg)*(9.8 m/s^2), so can you find the spring constant?

 

[flyingpig]

Oh lol

mg = kx

mg/x = (5*10)(3) = 50/3 = 16.7N/m

16.7N/m * -3.5 * 10^-4 + 50 * 1 * 10^-2 = 0.49J Still positive?