Hooks Law F = kx Book answer wrong I think

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In summary, the book says that a spring with a constant of 0.2 N/cm can stretch a distance of 5 cm, but the answer provided is 1 Newton.
  • #1
Barclay
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1

Homework Statement


[/B]
Hello, two basic questions on the same topic and I think the book answer is wrong so I've posted both here if that's okay.

Q1. What force is required to stretch a spring with spring constant 0.2 N/cm a distance of 5 cm?

Q2. A vertical spring stretches 5 cm under a load of 100 g. Determine the spring constant.


2. Homework Equations

F = kx k = spring contant x = extension in spring

The Attempt at a Solution



Q1. F = kx
F = 0.2 x 5
F = 1 N

1 Newton is my answer but the book says 0.01 N.

Q2. 100g = 0.1kg therefore weight (force) = 0.1 x 10 = 1 N

F = kx
1 = 5k
k = 0.2 N/cm

0.2 N/cm is my answer but the book says 0.05 N/cm. I can't see where I've gone wrong. Please advise.
 
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  • #2
Barclay said:
Q1. What force is required to stretch a spring with spring constant 0.2 N/cm a distance of 5 cm?

Are you sure the dimension of the spring constant is N/cm and not N/m?

Barclay said:
Q2. A vertical spring stretches 5 cm under a load of 100 g. Determine the spring constant.

The solution in the book seems to be the inversed value of the spring constant, again using the common dimension N/m for it.
 
  • #3
stockzahn said:
Are you sure the dimension of the spring constant is N/cm and not N/m?

The solution in the book seems to be the inversed value of the spring constant, again using the common dimension N/m for it.

Yes absolutely sure
 
  • #4
Barclay said:
Yes absolutely sure

Your calculations seem to be correct to me.

Q1:

If you take k = 0.2 N/m, then

F = 0.2 N/m ⋅ 0.05 m = 0.01 N

Q2:

If you take k in N/m, then

k = 1 N / 0.05 m = 20 N/m → 1/k = 0.05 m/N

Maybe this is where the answers are coming from, but they have to have messed up several times.
 
  • #5
1 Newton is equal to 1 kg m/s2. The cm and m cannot cancel each other out. Change the cm to m first, then solve.
 
  • #6
Barclay said:
Q1: 1 Newton is my answer but the book says 0.01 N.
1N is correct unless somebody meant k = 0.2N/m.
Q2: 0.2 N/cm is my answer but the book says 0.05 N/cm.
The book here is wrong no matter what. Your answer is correct.
 
  • #7
Thanks all. The book authors were trying to explain (somewhere earlier in the text) that the "spring constant can be in N/cm or N/mm as long as the extension is measured in the same units". So I think the authors go mixed up with all the units somewhere because they've been working in metres up to that point in the book
 
  • #8
dmcdaniel5253 said:
1 Newton is equal to 1 kg m/s2. The cm and m cannot cancel each other out. Change the cm to m first, then solve.

I just read this and seems to make sense.
So for Q1 What force is required to stretch a spring with spring constant 0.2 N/cm a distance of 5 cm?

0.2N/cm = 20N/m
5 cm = 0.05 m

F = kx
F = 20 x 0.05
F = 1 N

So the same answer as mine. This book is just wrong.
 
  • #9
Barclay said:
Thanks all. The book authors were trying to explain (somewhere earlier in the text) that the "spring constant can be in N/cm or N/mm as long as the extension is measured in the same units". So I think the authors go mixed up with all the units somewhere because they've been working in metres up to that point in the book
Ya. get a new book!
 

Related to Hooks Law F = kx Book answer wrong I think

1. What is Hooke's Law?

Hooke's Law is a principle in physics that states the force required to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed.

2. What does the equation F = kx represent in Hooke's Law?

The equation F = kx represents the relationship between force (F), spring constant (k), and displacement (x) in Hooke's Law. It shows that the force applied to a spring is directly proportional to the displacement of the spring.

3. How is Hooke's Law applicable in real-life situations?

Hooke's Law can be applied in many real-life situations, such as in the design of spring-based objects like mattresses, car suspension systems, and trampolines. It is also used in the calibration of instruments like weighing scales and force gauges.

4. What is the significance of the spring constant (k) in Hooke's Law?

The spring constant (k) is a measure of the stiffness of a spring. It is a constant value that is unique to each spring and determines how much force is required to stretch or compress the spring by a certain distance. A higher spring constant indicates a stiffer spring, while a lower spring constant indicates a more flexible spring.

5. Can Hooke's Law be applied to materials other than springs?

Yes, Hooke's Law can be applied to some materials other than springs, such as rubber bands and elastic materials. However, it is important to note that it is only applicable within the elastic limit of a material, meaning that if the force applied exceeds a certain point, the material will no longer follow Hooke's Law and can become permanently deformed.

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