Force Constant k of Spring - Potential Energy Homework

AI Thread Summary
The discussion revolves around calculating the force constant k of a spring used to launch a 2 kg rock to a height of 500 m after being compressed 1 m. The potential energy (P.E.) of the rock is calculated using the formula P.E. = mgh, resulting in 10,000 joules when corrected for SI units. The elastic potential energy of the spring is expressed as P.E. elastic = (1/2)(k)(x^2), leading to the calculation of k as 20,000 N/m. Participants emphasize the importance of using correct SI units throughout the calculations. The final conclusion confirms that the calculated spring constant indicates a very strong spring.
themadhatter1
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Homework Statement


A spring is compressed a distance of 1m and a 2 kg rock is shot straight up with the spring. The rock attains a maximum height of 500m. What is the force constant k of the spring?

Use g=10m/s^2

Homework Equations


P.E.= mgh
P.E. elastic= (1/2)(k)(x^2)

The Attempt at a Solution



P.E.= mgh
P.E.=(2000g)(10m/s^2)(500m)
P.E.= 10000000 Newtons

P.E. elastic= (1/2)(k)(x^2)
k= P.E elastic/ (1/2)(x^2)
k= 10000000/ (1/2)(1)
k=2000000 Newtons ?
 
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themadhatter1 said:
P.E.= mgh
P.E.=(2000g)(10m/s^2)(500m)
P.E.= 10000000 Newtons
Use SI units ie kilogram, not gram. Furthermore, energy is in joules, not Newtons! Also, the spring constant k is in Newtons per meter, not Newtons. Please read up on units, okay?
 
Fightfish said:
Use SI units ie kilogram, not gram. Furthermore, energy is in joules, not Newtons! Also, the spring constant k is in Newtons per meter, not Newtons. Please read up on units, okay?

Thanks,

So, if I correct my units I would have:

P.E.= mgh
P.E.=(2kg)(10m/s^2)(500m)
P.E.= 10000 joules

P.E. elastic= (1/2)(k)(x^2)
k= P.E elastic/ (1/2)(x^2)
k= 10000/ (1/2)(1)
k=20000 N/m ?

Is this correct?
 
Looks good; that is one very strong spring indeed.
 
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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