Hooke's Law- measurement of mass/weight

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Homework Help Overview

The discussion revolves around applying Hooke's Law to determine the mass and weight of a body hanging from a spring. The spring constant is given, and the extension of the spring is measured.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between spring constant, mass, and weight using Hooke's Law. There are attempts to rearrange equations to solve for mass and weight, with some questioning the validity of their calculations and units.

Discussion Status

There is ongoing exploration of the correct application of formulas and units. Some participants have provided guidance on rearranging equations and checking units, while others are still clarifying their understanding of the relationships involved.

Contextual Notes

Participants are discussing the implications of their calculations and the importance of unit consistency. There is a recognition of the need to verify answers by substituting back into the original equations.

thennathalie
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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
 
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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.
 
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?
 
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?
 
My unit for weight is N.
And for Mass is Kg.

I have not, didn't think about that. Thanks
 
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
 
Thanks for keeping up with me.
I am new to physics.

That means i would have to multiply 104*50 = 5200?
 
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?
 
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:
 
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  • #11
☺️ thank you so much.
 

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