Calculating Weight of a Brick Using Laws of Motion: 42N, 52N, and Acceleration

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Homework Help Overview

The discussion revolves around calculating the weight of a brick using the laws of motion, specifically focusing on the relationship between applied force, acceleration, and weight. The original poster presents a scenario where the brick is lifted with two different forces, 42N and 52N, leading to different accelerations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of Newton's laws to relate applied forces and acceleration to the weight of the brick. There are attempts to derive equations for weight and acceleration based on the forces applied. Questions arise about isolating variables and the validity of the equations derived.

Discussion Status

Participants are actively engaging with the problem, with some providing hints and guidance on how to manipulate the equations. There is recognition of potential mistakes in the initial equations, and a shift towards considering net forces is noted. The discussion is ongoing, with various interpretations and approaches being explored.

Contextual Notes

Participants mention the need to account for net forces when applying Newton's second law, indicating a focus on understanding the setup of the problem. There is also mention of the original poster's uncertainty regarding the manipulation of equations and the cancellation of terms.

jen333
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Using Laws of Motion!

Homework Statement


The acceleration of a brick when lifted by 42N is a. The acceleration of the same brick lifted by 52N is 2a. What is the weight of the brick?


Homework Equations


Fg=mg
Fapp=ma


The Attempt at a Solution


I don't really know where to start, here's what I've done so far:
W=mg
Fapp=ma, m=F/a

W=(F/a)g
W=((42N)(9.81m/s[tex]^{2}[/tex]))/a

W((52N)(9.81m/s[tex]^{2}[/tex]))/2a

hopefully that's a good start. thanks
 
Last edited:
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Thats a great start!

So, you have two equations and two unknowns. A good rule to keep in mind is that if you have the same number of equations and unknowns, then it is possible to find both unknowns. Thus, you should be able to solve for W, and a, if you wish to.

HINT: You have to use the two equations for W you derived to get rid of a. Can you find a in terms of W? If so, you should be able to plug that in for a in one of those equations.
 
Thx Go1. Glad to know I'm on the right track.
I was wondering if I could get some clarification on getting rid of a...
if I can use my two W equations to isolate for a
I have:
a=(42N)(9.81)/W
2a=(52N)(9.81)/W

or in general: a=Fg/W

I attempted to plug in the first 'a' equation into the second:

2(42N)(9.81)/W=(52N)(9.81)/W

buut, i don't think that's the way to do it since W cancels out...h'mm..
 
Last edited:
for some reason my edit button isn't working...
but, to just show what I mean with the W equations:

824.04/w=510.12/w

I'm sorry, but I think that i just might've understood your question wrong.
 
OK I think I see why the weight is canceling out above. The equations we have aren't correct. (I missed the mistake as well.)

Think of it this way, when a block is lifted by 42N:

[tex]\Sigma F=ma[/tex]

But, [tex]\Sigma F= 42N-W[/tex]

Set up the equations from Newton's second law remembering that we need the NET force on the block as it is lifted up, not just the applied force. See if you can get an answer going about the problem this way.
 
ok,
so now to confirm...
ma=F=42N-W
and
m(2a)=2F=52N-W

2(42N-W)=52N-W
84N-2W=52N-W
W=32N

?
 
Last edited:

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