Hooke's Law and the force constant

[mandy9008]

Homework Statement

When a 2.90-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.93 cm.
(a) What is the force constant of the spring?
(b) If the 2.90-kg object is removed, how far will the spring stretch if a 1.45-kg block is hung on it?

Homework Equations

Hooke's law: Fs=-kx
F=ma

The Attempt at a Solution

a. F=ma
F= (2.90 kg) (9.8 m/s²)
F=28.42N

Fs=-kx
27.42 N=-k (0.0293m)
k=970.0 N/m

b. F=ma
F= (1.45 kg) (9.8 m/s²)
F=14.21 N

Fs=-kx
14.21 N = (970 N/m) x
x=0.015 cm

 

[collinsmark]

a. F=ma
F= (2.90 kg) (9.8 m/s²)
F=28.42N

Fs=-kx
27.42 N=-k (0.0293m)
k=970.0 N/m

b. F=ma
F= (1.45 kg) (9.8 m/s²)
F=14.21 N

Looks okay to me, so far.

Fs=-kx
14.21 N = (970 N/m) x
x=0.015 cm

I think you might have made a mistake in your meters to centimeters conversion.

 

[pgardn]

Homework Statement

When a 2.90-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.93 cm.
(a) What is the force constant of the spring?
(b) If the 2.90-kg object is removed, how far will the spring stretch if a 1.45-kg block is hung on it?

Homework Equations

Hooke's law: Fs=-kx
F=ma

The Attempt at a Solution

a. F=ma
F= (2.90 kg) (9.8 m/s²)
F=28.42N

Fs=-kx
27.42 N=-k (0.0293m)
k=970.0 N/m

b. F=ma
F= (1.45 kg) (9.8 m/s²)
F=14.21 N

Fs=-kx
14.21 N = (970 N/m) x
x=0.015 cm

Are the numbers in red supposed to be the same?

 

[mandy9008]

Are the numbers in red supposed to be the same?

Yes, the numbers are the same, just a typo.

My answer was just off by a multiple of 10, the final answer was 1.5 cm.

I have another question pertaining to this also.

How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
Do I use the two x values to find delta x and solve for work using W=Fx?

 

[collinsmark]

How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?
Do I use the two x values to find delta x and solve for work using W=Fx?

No, W=Fx does not apply to this problem. W=Fx only applies if F is constant. But the force on a spring is most certainly not constant as you change x.

However, your textbook should have an equation for the potential energy stored by a spring. How does the work-energy theorem fit in once you find that?

[Edit: By the way, if you happen to wonder why W=Fx doesn't apply to this problem, it takes a bit of calculus. If you don't know calculus, you can ignore this paragraph. While W=Fx is true for a constant force, the general relationship is found by the line integral, W = \int \vec F \cdot \vec {ds}. If F is constant it can be pulled out from under the integral giving the result W=Fx. But for the case of a spring, F = kx is not a constant force. So the work equation is W = k\int x dx. Although I suggested that you look up the equation in your textbook, if you know calculus, you can now derive the equation yourself by evaluating the above integral (it is a definite integral -- I just left off the limits for brevity).]

 

[mandy9008]

the equation that I found was PE=1/2kx²
I found the new k value by plugging in the new number
F= (2.90 kg) (9.8 m/s2)
F=28.42N

Fs=-kx
28.42 N=-k (0.0750 m)
k= 378.9 N/m

PE=1/2 (378.9 N/m) (0.075m)²
PE=1.066J

 

[collinsmark]

Hello mandy9008,

the equation that I found was PE=1/2kx²
I found the new k value by plugging in the new number
F= (2.90 kg) (9.8 m/s2)
F=28.42N

Fs=-kx
28.42 N=-k (0.0750 m)
k= 378.9 N/m

PE=1/2 (378.9 N/m) (0.075m)²
PE=1.066J

Sorry, that's not the right approach. The only way to change the spring constant of a single spring, is to essentially damage the spring. You could stretch it apart so far that it goes through plastic deformation, such that it no longer reaches equilibrium at the same distance. Or you could very tightly wrap it around a different sized pipe, again forcing plastic deformation. Or you could cut a chunk off of it. Or you could melt it down and roll it into a brand new spring of a different shape/size. But that's not the point of this problem. Your new problem statement specifies, "...to stretch the same spring..." If it's the same spring, it implies the same spring constant.

So you already know the spring constant.

You found the correct equation for the potential energy of a spring.

P.E = (1/2)kx²

Invoking conservation of energy, that's the amount of work that needs to be put into the spring to stretch it by a distance x. You already know k from an earlier part of the problem, so it's just a matter of plugging in the numbers.

===========================
On a related tangent, you still might be wondering why W = Fx for a constant force, and why W =(1/2)kx² for a spring. I've thought of a way to explain this without calculus.

If you plot force F verses distance x, the work W is the area under the curve. I suggest doing this. Having an understanding of this relationship can really help in the future.

Start with a constant force F.
Plot force vs. distance. The y-axis is force. The x-axis is distance. Since the force is constant, the plot is simply a horizontal line, starting at 0, and ending at some final distance x. In other words, the area should look like a rectangle. One side (the vertical side) of the rectangle has a height F. The other side (the horizontal side) has a length x. Find the area under the curve, and call it W. (Hint: you should end up with W = Fx).

Now move on to a different plot of the spring force F = kx.
The plot should start at the origin, and move up in a diagonal line as x increases, finally ending at some final x along the x-axis and some F = kx along the y-axis (k is the slope of the line). The area under the curve should look like a right triangle. The horizontal side has length x. The vertical side has height kx. Find the area of the triangle, and call it W. (Hint: you should end up with W = (1/2)kx²).