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~christina~
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[SOLVED] Spring and Hook's law
spring question
spring question
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Doc Al said: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]
Doc Al said:Yes, 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.~christina~ said: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]
Doc Al said: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.
Beats me. Perhaps they want you to find the speed using Newton's 2nd law directly. (Which ends up being the same thing.)
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.
Spring and Hook's law is a scientific principle that describes the relationship between the force applied to a spring and the resulting displacement or extension of the spring.
The formula for Spring and Hook's law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement or extension of the spring.
Hook's law applies to real-life situations, such as the behavior of springs and elastic materials, the movement of pendulums, and the stretching of rubber bands.
The spring constant, represented by the letter k, is a measure of the stiffness of a spring. It is a constant value that determines the amount of force needed to extend or compress a spring by a certain distance.
The unit of measurement for the spring constant depends on the system of units being used. In the SI system, the unit for the spring constant is Newtons per meter (N/m).