Calculating Tension in a Falling 5kg Brick on a Rope | Homework Problem

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To find the tension in a rope holding a 5kg brick accelerating downward at 1.5 m/s², the equation T = mg + ma is used, where mg equals the weight of the brick (49 N). The calculation shows T = 49 N + 5.0 N(-1.5) = 41.5 N. The discussion highlights confusion over the correctness of the solution, with participants confirming the equation's validity despite initial doubts. The key takeaway is understanding the application of forces and the role of acceleration in calculating tension. This problem illustrates the importance of correctly applying Newton's second law in dynamics.
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Homework Statement


A 5kg brick on a rope is lowered towards the ground. While the brick is accelerating downward at 1.5m/s^2, find the tension in the rope. What is the reaction force to the upward force of tension acting on the brick?

Homework Equations


Sum of F = ma
w = Fg = mg
g = 9.8 m/s^2

The Attempt at a Solution


So basically what I did was drew a diagram.

/\ T
#
#
# a = -9,8 m/s^2
#
#
\/ W = mg = 5.0(9.8) = 49 N

T = mg + ma = 49 + 5.0(-1.5) = 41.5 N

O.k. where did i go wrong? lol I can't figure this out and it's driving me crazy
 
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Doesn't look wrong to me...why do you think it is wrong?
 
Because I always do something wrong...You sure it's correct? I'm just really shocked I got it correct it seems to easy i must be missing something.
 
You noted correctly that T=mg + ma, where a is negative. How did you arrive at that equation? That's the most important part.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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