Recent content by Hannah

so just one thing since i used conservation of energy my system would have to isolated so would i have to include the Earth as my system or just track+spring+car since its isolated until it reaches the top part of the track?

okay thank you very much for all your help ! :)

if i do it my original way i get k =2699.45 which is even bigger

0+1/2k(0.02)^2+mg(0)=1/2(0.1)(3.286)^2+1/2(m)(0)^2+(0.1)(-9.8)(0.5) 0.0002k=0.53989-0.49 k=249.45 seems pretty large

there is supposed to be an equal sign in the middle there instead of +, right after (-0.02)^2

1/2(0.1)(0)^2+1/2k(-0.02m)^2+1/2(0.1)(3.286)^2+1/2k(0)^2 wouldn't in the final case the spring be back at equilibrium so x=0 and the velocity after is 3.286 which i determined earlier

1/2(0.1)(0)^2 +mg(0)+1/2k(-2)^2=1/2(0.1)(3.286)^2+mg(0)+1/2(k)(0)+Ehf is that correct so far would the forces of gravity be 0 since there is no height and would the thermal energy be 0

yes, i think we just may have used something else for potential energy without the spring constant which we just learned not too long ago

well in an isolated system energy is conserved, i found this equation online but i have never used it in class before 1/2mvi^2+mgh,i+1/2kxi^2=1/2mvf^2+mghf+1/2kxf^2+EHf

con so it says conservation of energy but I am still confused on how that would me find acceleration? sorry I'm so bad at this lol

okay so normal force would be -mg to counteract the force of gravity on the object = -(0.1kg)(9.8m/s^2)=0.98

well if we know the final velocity at the top we could figure out the initial at the bottom by vf^2=vi^2 +2ad so Vi=3.286 but to get the acceleration at the bottom we would need either time or distance .

So is it zero? I was thinking that before but I didn't think it made sense because then kinetic force would be 0.

No just in the horizontal direction it has negative acceleration.

Ya i was starting to rethink that part I think since the spring accelerates the car from 0m/s to 1m/s Fsc,x=mg+ma