~christina~
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[SOLVED] Spring and Hook's law
spring question
spring question
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so then I guess the negative will cancel out then... is it supposed to be like this then?You left out the minus sign in computing the potential energy function (which is why you get strange answers in part b):
[tex]U = - \int F_x(x) dx[/tex]
ok. well now with the corrected eqznYes, but I'd write it this way:
[tex]U = - \int F_x (x) dx = -\int (-60.0N/m)x + (-18.0N/m^2) x^2 dx = (30.0N/m) x^2 + (6.0N/m^2) x^3 + C [/tex]
Since U = 0 for x = 0, the constant of integration (C) is zero.
This equation has spring PE terms for a Hooke law spring, not the one in this problem. But you used the correct equation when you did your calculation.to get that would I just use...
[tex]K_i + U_{si} = K_f + U_{sf} = .5 mv_i^2 + .5 mx_i^2 = .5 mv_f^2 + .5 mx_f^2 [/tex]
Looks good to me.Hm..I figured out that I did something incorrect in the final equation..I didn't include the whole equation at all... thus the new equation would be...
[tex] 0.5(0.900kg)(0m/s)^2 + (30.0N/m)(1.00m)^2 + (6.00N/m^2)(1.00m)^3 = 0.5(0.900kg)(v_f)^2 + (30.0N/m)(0.500m)^2 + (6.00N/m^2)(0.500m)^3[/tex]
[tex] v_f= \sqrt{27.75 / (0.5*0.900)} = 7.85m/s [/tex]
hm...not sure if that's right but if my math is correct it should be..
Beats me. Perhaps they want you to find the speed using Newton's 2nd law directly. (Which ends up being the same thing.)c.) Use Newtonian dynamics to find the speed at this position
what exactly do they mean by "Newtonian dynamics" ?
Good.d.) Waht is the instantaneous power when x= 0.500m?
hm since I found the velocity
[tex] P_{average} = dW/ dt = F*v[/tex]
hm..would I think I'd use the original force equation and then plug in x= 0.500m then multiply it by the velocity that I found for that point assuming my velocity I found is correct of course.
Looks good. Don't get hung up on the minus sign in the force equation. That just means the force acts to the left. But the velocity also acts to the left, so (using the same sign convention) it should also be negative.[tex]F_x (0.500m)= -60.0N/m (0.500m)- 18.0N/m^2 (0.500m)^2 = -34.5N [/tex]
then I'd multiply that by [tex]v_{0.500}= 7.85m/s[/tex]
[tex] P_{average} = dW/ dt = F*v= (34.5N)(7.85m/s)= 270.825 W [/tex]
okThis equation has spring PE terms for a Hooke law spring, not the one in this problem. But you used the correct equation when you did your calculation.
F= ma ?Beats me. Perhaps they want you to find the speed using Newton's 2nd law directly. (Which ends up being the same thing.)
oh...yup I was confused about that...Looks good. Don't get hung up on the minus sign in the force equation. That just means the force acts to the left. But the velocity also acts to the left, so (using the same sign convention) it should also be negative.