Understanding the Relationship Between Spring and Hook's Law

[~christina~]

[SOLVED] Spring and Hook's law

spring question

 

[Doc Al]

Potential Energy Function

Read this to learn how to find the potential energy function given the force: http://hyperphysics.phy-astr.gsu.edu/hbase/pegrav.html#pef"

 

[~christina~]

I'm not sure about the negative though..is that correct? (way I integrated it)

a) calculate the potential energy function U(x)

I'm not sure what they mean by calculate but I assume they meant integrate and what I got was..

If I read correctly on that site that you linked me...the $$\int F_x(x)= U_x$$

$$F_x(x)= -\alpha x - \beta x^2$$

$$\alpha = 60.0 N/m$$
$$\beta = 18.0 N/ m^2$$

$$F_x(x)= - (60.0N/m)x - (18.0 N/m^2) x^2$$

$$\int F_x (x)= -\int (60.0N/m)x + (18.0N/m^2) x^2 =<br /> -(30.0N/m) x^2 - (6.0N/m^2) x^3$$

I'm not sure about the negative though..is that correct? (way I integrated it)

b) An object with mass m= 0.900kg on frictionless horizontal surface is atttatched to the spring and pulled a distance 1.00m to the right (+ x-direction) to the spring released . What is the speed of the object when it is 0.500m to the right of the equillibrium position?

Hm...

m= 0.900kg
$$x_i= 1.00m$$
$$x_f= 0.500m$$
$$v_i = 0m/s$$
$$v_f= ?m/s$$
to get that would I just use...

$$K_i + U_{si} = K_f + U_{sf} = .5 mv_i^2 + .5 mx_i^2 = .5 mv_f^2 + .5 mx_f^2$$

however I have U(x) ...I'm not sure how to incorperate that into the equation.
I would think that I replace the U in the equation with the equation to get...
with the numbers plugged in...

$$.5(0.900kg)(0m/s)^2 + (-30.0N/m)(1.00m)^2 = .5(0.900kg)(v_f)^2 + (-30.0N/m)(0.500m)^2$$

$$v_f=\sqrt{-22.5/ .45}$$

this would give a negative under the square root and the problem says to take the right as the positive direction so I'm thinking I should change something to a positive and thus I wouldn't get a negative but I'm not sure where that would be.
Would it be where the - is in the equation I'm using?

 

[Doc Al]

missing minus sign

You left out the minus sign in computing the potential energy function (which is why you get strange answers in part b):

$$U = - \int F_x(x) dx$$

 

[~christina~]

You left out the minus sign in computing the potential energy function (which is why you get strange answers in part b):

$$U = - \int F_x(x) dx$$

so then I guess the negative will cancel out then... is it supposed to be like this then?

$$\int F_x (x)= -\int - (60.0N/m)x - (18.0N/m^2) x^2 =<br /> (30.0N/m) x^2 + (6.0N/m^2) x^3$$

thanks

 

[Doc Al]

Yes, but I'd write it this way:

$$U = - \int F_x (x) dx = -\int (-60.0N/m)x + (-18.0N/m^2) x^2 dx = (30.0N/m) x^2 + (6.0N/m^2) x^3 + C$$

Since U = 0 for x = 0, the constant of integration (C) is zero.

 

[~christina~]

Yes, but I'd write it this way:

$$U = - \int F_x (x) dx = -\int (-60.0N/m)x + (-18.0N/m^2) x^2 dx = (30.0N/m) x^2 + (6.0N/m^2) x^3 + C$$

Since U = 0 for x = 0, the constant of integration (C) is zero.

ok. well now with the corrected eqzn

b) An object with mass m= 0.900kg on frictionless horizontal surface is atttatched to the spring and pulled a distance 1.00m to the right (+ x-direction) to the spring released . What is the speed of the object when it is 0.500m to the right of the equillibrium position?

Hm...

m= 0.900kg
$$x_i= 1.00m$$
$$x_f= 0.500m$$
$$v_i = 0m/s$$
$$v_f= ?m/s$$
to get that would I just use...

$$K_i + U_{si} = K_f + U_{sf} = .5 mv_i^2 + .5 kx_i^2 = .5 mv_f^2 + .5 kx_f^2$$

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...

$$0.5(0.900kg)(0m/s)^2 + (30.0N/m)(1.00m)^2 + (6.00N/m^2)(1.00m)^3 = 0.5(0.900kg)(v_f)^2 + (30.0N/m)(0.500m)^2 + (6.00N/m^2)(0.500m)^3$$

$$v_f= \sqrt{27.75 / (0.5*0.900)} = 7.85m/s$$

hm...not sure if that's right but if my math is correct it should be..


c.) Use Newtonian dynamics to find the speed at this position

what exactly do they mean by "Newtonian dynamics" ?

d.) What is the instantaneous power when x= 0.500m?

hm since I found the velocity

$$P_{average} = dW/ dt = F*v$$

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.

$$F_x (0.500m)= -60.0N/m (0.500m)- 18.0N/m^2 (0.500m)^2 = -34.5N$$

then I'd multiply that by $$v_{0.500}= 7.85m/s$$

$$P_{average} = dW/ dt = F*v= (34.5N)(7.85m/s)= 270.825 W$$


I think that's it except I don't know how to do c.)

Thanks

 

[Doc Al]

to get that would I just use...

$$K_i + U_{si} = K_f + U_{sf} = .5 mv_i^2 + .5 mx_i^2 = .5 mv_f^2 + .5 mx_f^2$$

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.

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...

$$0.5(0.900kg)(0m/s)^2 + (30.0N/m)(1.00m)^2 + (6.00N/m^2)(1.00m)^3 = 0.5(0.900kg)(v_f)^2 + (30.0N/m)(0.500m)^2 + (6.00N/m^2)(0.500m)^3$$

$$v_f= \sqrt{27.75 / (0.5*0.900)} = 7.85m/s$$

hm...not sure if that's right but if my math is correct it should be..

Looks good to me.

c.) Use Newtonian dynamics to find the speed at this position

what exactly do they mean by "Newtonian dynamics" ?

Beats me. Perhaps they want you to find the speed using Newton's 2nd law directly. (Which ends up being the same thing.)

d.) what is the instantaneous power when x= 0.500m?

hm since I found the velocity

$$P_{average} = dW/ dt = F*v$$

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.

Good.

$$F_x (0.500m)= -60.0N/m (0.500m)- 18.0N/m^2 (0.500m)^2 = -34.5N$$

then I'd multiply that by $$v_{0.500}= 7.85m/s$$

$$P_{average} = dW/ dt = F*v= (34.5N)(7.85m/s)= 270.825 W$$

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.

 

[~christina~]

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.

ok

Beats me. Perhaps they want you to find the speed using Newton's 2nd law directly. (Which ends up being the same thing.)

F= ma ?
I know that

F= ma= m(v/t)

t(F/m)= v

but I don't have the time...so I really don't get how would I find the velocity using F= ma

I don't exactly have the time though...how would I go about finding the velocity?

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.

oh...yup I was confused about that...


Thanks a lot

 

[~christina~]

Um..well since you didn't say anything about how to find the time...

since in F= ma

I have
d
v

but no t

v= d/t

t= d/v

I thought that I use that but

if I was trying to find the v from just the information given would I be able to find it?

In the beginning all I had was the distance, force, and potential energy...

how would I get velocity from that since I don't have the time or acceration?

 

[Doc Al]

Looks like this same problem was recently posted. Check it out: https://www.physicsforums.com/showthread.php?t=199147

The only thing I can think of doing for part c is exactly what andrevdh suggests, which ends up being equivalent to using energy methods.

 

[~christina~]

oh...how interesting...

THANKS FOR ALL YOUR HELP Doc Al