Spring-launched rollercoaster problem

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To design a spring-launched roller coaster, the spring constant must be calculated to ensure the car can ascend a 10m hill after descending 15m. The user is working on determining the spring constant (k) and maximum speed for a 400 kg car, with the spring compressed by 2.0m. The initial energy equation provided is correct, but the interpretation needs adjustment; the velocity is zero when the spring is fully compressed. After solving for the minimum k, a 10% safety factor should be added to ensure safety. The discussion emphasizes the importance of correctly applying energy conservation principles in roller coaster design.
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49. You have been hired to design a spring launched roller coaster that will carry two passengers per car. Te car goes up a 10m high hill, then descends 15m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.0 m and that a loaded car will have a maximum mass of 400 kg. For safety reasons, the spring constant should be 10% larger than the minimum needed for the car to just make it over the top.

a. What spring constant should you specify?
b. What is the maximum speed of a 350 kg car if the spring is compressed the full amount?

I'm having trouble with part a. Here's what I've got so far:

known:
m=400kg

y0= 5m
y1= 15m
y2= 0m

v1= 0 m/s

s0= -2.0m (compression of spring)
s1= 0m

to find:
k
v0

.5mv0^2+mgy0+.5kS0^2 = .5mv1^2+mgy1+.5kS1^2

200v0^2+19600-2k = 58800

am I on the right track? how do I find out what I need to get to solve for part a of the problem?
 
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Your equation is OK if it is interpreted correctly. On the left hand side the velocity is zero when the spring is compressed 2 meters. (There is no minus sign in front of the 2k in the next line.) On the right hand side, the spring is not compressed and if the minimum (no safety factor) k is used, the velocity is zero. Only the gravitational potential energy term survives. Solve for k and then add the 10% safety factor.
 
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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