Calculating Work Done by a Spring: A Simple Solution

AI Thread Summary
The discussion focuses on calculating the work done by a spring with a force constant of 2.5 x 10^3 N/m when stretched. The correct formula used is W=1/2kx^2, where 'k' is the spring constant and 'x' is the displacement. The work done in stretching the spring by 6.0 cm is confirmed to be 4.5 J, while the additional work for a further stretch of 2.0 cm is 3.5 J. A participant realizes they miscalculated due to using incorrect units for length. The conversation highlights the importance of unit consistency in physics calculations.
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[SOLVED] Easy Spring Constant Question

A particular spring has a force constant of 2.5 x 10^3 N/m. (a) How much work is done in stretching the relaxed spring by 6.0 cm? (b) How much more work is done in stretching the spring an additional 2.0 cm?



I am using W=1/2kx^2 but I am not completely sure.



Well the answer are a. 4.5 J and b. 3.5 J. When I plug in I get W=1/2(2500)(8)^2. I am getting the answers 45000 and 35000. I think I am using the right equation I am just multiplying wrong or something..
 
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What is the standard SI unit for length?
 
Ah man! I can't believe I missed that. Thank you!
 
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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