Hooke's Law- measurement of mass/weight

In summary, for a body of unknown mass hanging from a spring and stretching it to 50 cm with a spring constant of 104 N/cm, the weight of the body is 5200 N and the mass is 530.6 kg. The relevant equations used were k = Mg/∆L and W = Mg.
  • #1
thennathalie
6
0
A body of unknown mass hangs from a spring and stretches it to 50 cm. If the spring constant is 104 N/cm determine:

a. the weight of the body
b. the mass of the body


k = mg/∆L
W= Mg


mass : k = mg/∆L 50/104= .4807692308
weight : W= Mg .4807692308*9.8=4.711538462
 
Physics news on Phys.org
  • #2
thennathalie said:
A body of unknown mass hangs from a spring and stretches it to 50 cm. If the spring constant is 104 N/cm determine:

a. the weight of the body
b. the mass of the body


k = mg/∆L
W= Mg


mass : k = mg/∆L 50/104= .4807692308
weight : W= Mg .4807692308*9.8=4.711538462

In your attempt to find the mass you've calculated ΔL/k, which is not correct. Your Relevant Equation says k = Mg/ΔL. Rearrange that equation to solve for M. Note that there's a g in there!

Since Mg is the weight and Mg appears in your first Relevant Equation, you can find the weight first by rearranging the equation to place Mg on the left.
 
  • #3
Hi, and welcome to the Physics Forums.

What are the units of your answers. Have you substituted your answers back into the formulae as a check of their validity?
 
  • #4
Okay, so what you're telling me is that the equation will be Mg= k/∆L
and not k=Mg/∆L/.

using the new equation:

Mg= k/∆L
Mg= 104/50 which gives me 2.08
Mg= 2.08

so the weight would be 2.08?
 
  • #5
My unit for weight is N.
And for Mass is Kg.

I have not, didn't think about that. Thanks
 
  • #6
thennathalie said:
Okay, so what you're telling me is that the equation will be Mg= k/∆L
and not k=Mg/∆L/.

using the new equation:

Mg= k/∆L
Mg= 104/50 which gives me 2.08
Mg= 2.08

so the weight would be 2.08?

Ahhh. Nope :shy:

Hooks law is usually presented as (ignoring the sign implying the direction of the force):

F = k ΔL

Where F is the force applied to the spring, ΔL the change in its length.

In your case the force is a weight due to gravity, Mg, so that Mg = k ΔL

Your Relevant Equation said:

k = Mg/ΔL

This is just an algebraic rearrangement of the same formula. You need to be able to perform these manipulations of the formulas in order to be able to solve for whatever unknown values are buried in the formula
 
  • #7
Thanks for keeping up with me.
I am new to physics.

That means i would have to multiply 104*50 = 5200?
 
  • #8
thennathalie said:
Thanks for keeping up with me.
I am new to physics.

That means i would have to multiply 104*50 = 5200?

Yes, that's what the formula says. What are the units?
 
  • #9
Would be 5200 N because is weight.

So for the mass,

I would use the W=Mg formula.

5200=m9.8
M=530.6kg
 
  • #10
thennathalie said:
Would be 5200 N because is weight.

So for the mass,

I would use the W=Mg formula.

5200=m9.8
M=530.6kg

Yup. Looks good. :smile:
 
  • Like
Likes 1 person
  • #11
☺️ thank you so much.
 

1. What is Hooke's Law?

Hooke's Law is a physical law that states that the force needed to extend or compress a spring is directly proportional to the distance the spring is extended or compressed.

2. How is Hooke's Law used to measure mass/weight?

Hooke's Law can be used to measure mass/weight by attaching a spring to the object and measuring the distance the spring is extended. The force of the spring can then be calculated using Hooke's Law and converted to mass/weight using the acceleration due to gravity.

3. What is the unit of measurement for the force in Hooke's Law?

The unit of measurement for the force in Hooke's Law is Newtons (N).

4. How is Hooke's Law different from other laws of motion?

Hooke's Law specifically applies to the behavior of springs and elastic materials, whereas other laws of motion, such as Newton's Laws, apply to a wide range of objects and materials.

5. Are there any limitations to Hooke's Law?

Yes, Hooke's Law is only applicable to objects that behave elastically, meaning they return to their original shape after the force is removed. It also assumes that the force applied is within the elastic limit of the material.

Similar threads

  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
813
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
449
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Classical Physics
Replies
24
Views
726
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top