How Is Spring Potential Energy Calculated?

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The potential energy of a spring is calculated using the formula PE = 1/2 k x^2, where k is the spring constant and x is the displacement. In this case, a force of 800 N stretches the spring 0.200 m, leading to a spring constant of 4000 N/m. The potential energy when stretched to 0.200 m is 80.0 J, derived from the spring constant and the displacement. When compressed by 5.00 cm (0.050 m), the potential energy is calculated to be 5.00 J, demonstrating that potential energy varies with the square of the displacement. Thus, the potential energy of a spring is directly related to how much it is stretched or compressed.
erik-the-red
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A force of 800 m stretches a certain spring a distance of 0.200 m.

1. What is the potential energy of the spring when it is stretched a distance of 0.200 m?

Why is the answer 80.0 J instead of 160. J?

2. What is its potential energy when it is compressed a distance of 5.00 cm?

Why is the answer 5.00 J?
 
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How does the potential energy of a spring depend on how much it is stretched or compressed?

Note that it's not saying that a force of 800 N was uniformly applied over a distance of 0.200 m; it says that a force of 800 N will stretch the spring 0.200 m. (Use that fact to find the spring constant.)
 
Thank you so much!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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