F=kx (Hooke's Law) Problem

• Feb 6th 2009, 03:29 PM
s3a
F=kx (Hooke's Law) Problem
Can someone guide me through this problem:

2. A mass of 500 g stretches a spring 8.0 cm when it is attached to it.
What additional weight would you have to add to it so that the spring is stretched 10 cm?
__________________________________________________ __________________________

I tried plugging in numbers into the rule and that didn't lead me anywhere!
• Feb 6th 2009, 03:41 PM
skeeter
$F = kx$

$mg$ is weight ... force due to gravity

$mg = kx$

$500g = k \cdot 8$

let $m$ = the additional mass (in grams) required ...

$(500+m)g = k\cdot 10$

divide the 2nd equation by the first ...

$\frac{(500+m)g = k\cdot 10}{500g = k \cdot 8}$

$\frac{500+m}{500} = \frac{10}{8}$

solve the last equation for $m$
• Feb 7th 2009, 04:43 PM
Knowledge
Alternative Method
As stated, the equation for Hooke's Law can be rewritten as follows:

$F=ma=mg=-kx$

The gravitational constant $g=9.81\left(\frac{m}{s^{2}}\right)$ is substituted for acceleration $a$, $k$ represents the spring constant, and $x$ is the displacement (distance stretched in meters).

Rearrange in terms of $k$ and solve:

$-k=\frac{mg}{x}$

$-k=\frac{(0.50kg)(9.81\left(\frac{m}{s^{2}}\right)) }{0.08m}$

$k=-61.31\left(\frac{N}{m}\right)$

Now rewrite the equation in its standard form $F=-kx$ and evaluate using the new length.

The rest is pretty straight-forward, just remember that weight is a measure of gravitational force and mind your significant figures (2).
• Feb 7th 2009, 11:02 PM
Hooke's Law
Hello s3a
Quote:

Originally Posted by s3a
Can someone guide me through this problem:

2. A mass of 500 g stretches a spring 8.0 cm when it is attached to it.
What additional weight would you have to add to it so that the spring is stretched 10 cm?
__________________________________________________ __________________________

I tried plugging in numbers into the rule and that didn't lead me anywhere!

Forget formulae. Just use common sense. Hooke's Law says that:

• The tension in an elastic string is proportional to its extension.

In other words, if you double one, you double the other; if you increase one by 50%, you increase the other by 50%, and so on.

You want to increase the extension from 8 cm to 10 cm; in other words, by 25%. So increase the tension by 25% as well. Put on an extra mass of 25% of 500 gm. In other words, you'll need an extra 125 gm.

That was easy, wasn't it?

• Feb 8th 2009, 02:10 AM
elliotyang
Well, you can solve it using common sense. But using formula, and understand enable you to solve any question.
F=kx
mg=kx
500g=8k (1) initial condition
let say the weight to strecth the spring 10cm is M
Mg=10k (2)
(2)/(1):
M/500=10/8
you will get M = 625g
so it will require additional 125g
• Feb 8th 2009, 03:02 AM
Hooke's Law
Hello elliotyang

Of course, I'm not suggesting that you should always forget the formula. But you should always look for a simple approach first! When solving a quadratic, for example, always try to factorise before reaching for the formula.

• Feb 8th 2009, 03:12 PM
s3a
• Feb 8th 2009, 03:29 PM
skeeter
Quote:

Originally Posted by s3a

mass is not weight ... mg = W

125 g = .125 kg

(.125 kg)(9.8 m/s^2) = 1.225 N
• Feb 8th 2009, 03:50 PM
Knowledge
Quote:

Originally Posted by Knowledge
As stated, the equation for Hooke's Law can be rewritten as follows:

$F=ma=mg=-kx$

The gravitational constant $g=9.81\left(\frac{m}{s^{2}}\right)$ is substituted for acceleration $a$, $k$ represents the spring constant, and $x$ is the displacement (distance stretched in meters).

Rearrange in terms of $k$ and solve:

$-k=\frac{mg}{x}$

$-k=\frac{(0.50kg)(9.81\left(\frac{m}{s^{2}}\right)) }{0.08m}$

$k=-61.31\left(\frac{N}{m}\right)$

Now rewrite the equation in its standard form $F=-kx$ and evaluate using the new length.

The rest is pretty straight-forward, just remember that weight is a measure of gravitational force and mind your significant figures (2).

s3a,

If you had followed my methodology, then you would have arrived at the correct answer. The above process is most likely the preferred one for a high school physics student.

To conclude where I left off above:

$F=-kx$

$F=-(-61.31\left(\frac{N}{m}\right))(0.10m)$ Notice how the units cancel.

$F_(required)=6.131N$

$F_(original)=(0.50kg)(9.81\left(\frac{m}{s^{2}}\ri ght))=4.905N$

$F_(additional)=F_(required)-F_(original)$

$F_(additional)=6.131N-4.905N$

$F_(additional)=1.2N$
• Feb 8th 2009, 05:53 PM
elliotyang
Dear Knowledge,

I think your approach is kind of long winded.

For this case, F=-kx=mg

it is obviously seen that g and k are constant when compare the two situation since g is 9.81m/s^2 while k=spring constant. (since using the same spring, this vqlue should be the same for both)

So from this, we can deduce that m is directly proportional to x.
use this relation would be much faster.

instead you plug in the value into the formula compute, and then reverse it. of course it is not wrong, just kind of wasting time.
• Feb 8th 2009, 08:02 PM
Knowledge
Quote:

Originally Posted by elliotyang
Dear Knowledge,

I think your approach is kind of long winded.

For this case, F=-kx=mg

it is obviously seen that g and k are constant when compare the two situation since g is 9.81m/s^2 while k=spring constant. (since using the same spring, this vqlue should be the same for both)

So from this, we can deduce that m is directly proportional to x.
use this relation would be much faster.

instead you plug in the value into the formula compute, and then reverse it. of course it is not wrong, just kind of wasting time.

ElliotYang,

While my approach/explanation may be kind of "long winded" for people who have already had advanced math/physics classes, it is obvious the OP did not follow the rationale of using a proportion, nor did the OP understand the concept of Hooke's Law; i.e. the difference between mass and weight (hence the reason I included all of the units).

Furthermore, it is my opinion that this is not a forum to merely supply answers for a student's homework problems, but a place to further their understanding of the concept(s) and reinforce proper methods in preparation for higher level courses.

If the OP, or anyone for that matter, wants to be successful in advanced physics, it is a good idea to develop sound habits for problem solving from the beginning - because the concepts, equations, and methods only become more complex and involved.
• Feb 8th 2009, 08:16 PM
elliotyang
Quote:

it is a good idea to develop sound habits for problem solving from the beginning - because the concepts, equations, and methods only become more complex and involved.
I agree must understand the concept for problem solving. I just try to provide alternative, easier way to solve. Reduce it to the simplest form then only plug in the value. Too many value will sometimes lead to careless, besides need to take care about significant figures and decimal places all.
• Feb 22nd 2009, 01:51 PM
JahKnows
Easiest way using canadian teaching method
First easiest way to do it is to first get all your units in their easiest form in canada on the end of year exam they usually ask you use kg-m-N

so

500g becomes .5kg -- F=mg F=.5(9.8) = 4.9N
80cm becomes .08m

so now using the rule F=kx you need
Force in N=force constant N/m * displacement in m

4.9N = k * .08m
k=4.9/.o8
k=61.25

It is the same spring so same K value, we are now given a new displlacement

k=61.25 N/m
x=10cm = .1m

With this info we can find the Force

F=kx
F=(61.25)(10)
F=6.125N

Now that we have the new force you can get the total weight

F=mg
6.125=m(9.8)
m=.625

that is the total weight they asked you how much was added on so just take your total weight and substract your initial weight