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

AI Thread Summary
The discussion revolves around calculating the weight of a brick using Newton's laws of motion. The problem states that a brick experiences different accelerations when lifted with forces of 42N and 52N, leading to the equations W=mg and Fapp=ma. Participants suggest isolating acceleration (a) in terms of weight (W) to solve the equations simultaneously. Through manipulation of the equations, it is confirmed that the weight of the brick is 32N. The conversation emphasizes the importance of correctly applying Newton's second law to find the net force acting on the brick.
jen333
Messages
58
Reaction score
0
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^{2}))/a

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

hopefully that's a good start. thanks
 
Last edited:
Physics news on Phys.org
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:

\Sigma F=ma

But, \Sigma F= 42N-W

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:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top