Why Is the Tension in the Rope Not Matching the Book's Calculation?

[Gear2d]

Homework Statement

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?

Homework Equations

T=mg
F=ma

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,

Homework Statement

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?

Homework 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:

<br /> \sum \vec F = m \vec a \mbox{ or } \vec F_{\rm net} = m\vec a<br />

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

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

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.

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),