How Many Dynes to Change Length of Spring by 20cm?

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To change the length of a spring by 20 cm, it requires 400 dynes, calculated by multiplying the force needed for a 5 cm change by four. Springs operate linearly, allowing for direct proportionality in force and displacement. For a more formal approach, one can determine the spring constant (k) using the formula Fspring = k*x, where Fspring is the force and x is the displacement. The discussion confirms that the initial calculation is correct and no additional formulas are necessary. Understanding the linear nature of springs simplifies the calculation process.
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The question is... suppose it requires 100 dynes to change the length of a certain spring 5 cm. What force is required to change the length of the spring 20 cm?

Would you just multiply it... and say it would take 400 dynes?
 
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Sounds good to me :smile: Springs are linear. If you wanted to be more formal about it, you could find the spring constant k from Fspring=k*x since you know Fspring and you know x.
 
Okay, thank you! Just wanted to make sure i wasn't missing a formula somewhere
 
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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