Tension, mass, and velocity

[Gear2d]

1. The problem statement, all variables and given/known data


I have a 12kg block that is raised by a rope. If the velocity of the mass is decreasing at a rate of 5 m/s^2, what is tension in the rope?

2. Relevant equations

T=mg
F=ma

3. The attempt at a solution

My solution: T = mg+ ma = 180N

Book solution: T+ma = mg => 60N

I am confused as why you are subtracting here. I see that acceleration is in the downward direction (as stated by the question stem), but the object is still been raised. So shouldn't it be T = mg+ ma? Because to me, T+ma =mg looks like that acceleration of the mass is the upward direction (if that were the case the object would be increasing it speed not decreasing),

[alphysicist]

Hi Gear2d,

1. The problem statement, all variables and given/known data


I have a 12kg block that is raised by a rope. If the velocity of the mass is decreasing at a rate of 5 m/s^2, what is tension in the rope?

2. Relevant equations

T=mg

This equation is not true.


F=ma

I think you might need to be a bit more careful with this equation. This equation should be either:


\sum \vec F = m \vec a \mbox{ or } \vec F_{\rm net} = m\vec a


and when you actually use it here, for example in the y direction, you get:


\begin{align}
\sum F_y = m a_y\nonumber\\
F_{1y}+F_{2y} = m a_y\nonumber
\end{align}


since there are two forces. So what are the y-components, including sign, of the tension and weight forces? And what is the y-component of the acceleration? Those, with the correct sign, are what go into the force equation.



3. The attempt at a solution

My solution: T = mg+ ma = 180N

Book solution: T+ma = mg => 60N

I am confused as why you are subtracting here. I see that acceleration is in the downward direction (as stated by the question stem), but the object is still been raised. So shouldn't it be T = mg+ ma? Because to me, T+ma =mg looks like that acceleration of the mass is the upward direction (if that were the case the object would be increasing it speed not decreasing),