Recent content by andrewasd

Ok the baseline distance makes sense. I thought it was the relaxed height minus the start height. But I'm not sure what it equals then? And if it doesn't equal the velocity then why do that equation at all?

Right that's my mistake, it would = the final velocity, or v_f. I got 0.3 because I thought that the distance was the compressed distance, so since equilibrium is 0.7 minus compressed distance of 0.4 equals 0.3. And for the other one I thought it was the peak from equilibrium, so 1.2-0.7. Am...

Ok so in that case, the final equation would be the magnitude of KE(1/2 * 1.4 * 9.9 ^2) + PE_g(1.4 * 9.81 * 0.3) + PE_s(1/2 * 520.5 * 0.5^2) and the units would be m/s because its a velocity, is this correct?

Ok I think I'm starting to get it, the PE equation would be m*g*h I belive, and so the answer would the magnitude of TE. The equation your talking about it I think is U=1/2 k x^2, but am not completely sure, how do you then add it to the total energy though?

I'm not sure what youre getting at sorry, I thought it was one of the two equations I've mentioned. I'm confused. is it the 1/2*m*v^2 ?

I'm confused? I thought the most straightfoward way to do it would be Hookes law, F=-k*x but you say not to use forces? Is there an equation that I can get the velocity from directly? Maybe it's v=sqrt((k/m)*change in length ^2)

I'm guessing that we use the equation F(_spy) = -k*Change in length. The change in length is 0.8m because that is just 1.2-0.4. I also know the magnitude equation, which is the sqrt(first velocity ^2 + second velocity^2). I then drew it out on a graph to help visualize

Yes sorry, I am new to the forums. I've attempted with Hookes Law, and have attempted it but am confused

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