# Work by Tension for Mass Moving Down w/ Constant Acc.: WT

• iJamJL
In summary: I think I might have entered it wrong because the given information has two significant digits. The answer came out wrong according to my online homework. :cry: Normally it requires three or more significant digits, and so I just entered the full decimal. It gave it to me as... 274.848. I think I might have entered it wrong because the given information has two significant digits.
iJamJL

## Homework Statement

ASSUME: For all parts of this question, friction is negligible.

A vertical rope is used to lower a block of mass M = 40.9 kg at a constant acceleration of magnitude a = 3.09 m/s2. Find WT, the work done by the tension in the cord if the mass moves down distance s = 8.01 m.

W=F*s
F=mg
F=ma

## The Attempt at a Solution

My problem really roots from my inability to draw and set up the problem correctly because I never truly understood how to do that with tension.

Nevertheless, I do know that since the box is going down, and gravity goes in that direction as well, we could make tension positive. Since tension is positive, does that mean we set it up so that T-mg=ma?

The next part would be the box being lowered at a=3.09m/s/s. Since it's being lowered, which is the same direction as mg, does that mean that T is still positive? So we would have:

T-m(a1)=m(a2)
T-mg = m(a2)

From above. Obviously, something is wrong here. Which T should be negative? Also, I don't really understand which accelerations to use for the result. In T-m(a1), a1 is the acceleration that is given. I wasn't sure what to put after that, so I just put a2. Should it be 0? I'm just confused overall about setting this up.

Making a Free Body Diagram (FBD) of the forces acting on the block should help to dispel your confusion. Indicate your chosen axes so that directions will be clear. Next sketch in the velocity and net acceleration vectors. What directions do they have on your chosen axes?

gneill said:
Making a Free Body Diagram (FBD) of the forces acting on the block should help to dispel your confusion. Indicate your chosen axes so that directions will be clear. Next sketch in the velocity and net acceleration vectors. What directions do they have on your chosen axes?

Well, I did draw it. I have it so that the positive direction of y is up, meaning that the vertical rope is lowering the box in the negative direction, and gravity is also doing work in the negative direction. I want to say that I should set it up like this:

Work(T)= [(m*-a)+(-mg)]*h

Tension goes in the positive direction of the y-axis, and both the acceleration and gravity are negative. Therefore, I added them together to get the total force, and then multiplied it by the distance, which is h (Work = F*s). However, I am not sure whether the tension is equal to that. If it is, then the answer that I'm getting is:

Work(T)=[(40.9*-3.09)+(40.9*-9.81)]*8.01
=4226.1561 J
Edit: This might be the solution but I'm just giving it a shot..

Instead of having them in equal directions, we break it down so that:

-T=ma
T=-mg

We add them together, and get that 0=ma-mg. If we do that, then we can get F=m(a-g), which turns out to be -53.827. Then we go back to W=Fs, and we multiply 53.827*8.01 to get 431.15.

Last edited:

As you say, the acceleration g is negative as is the net acceleration a. Tension force is upwards. So:

T - M*g = -M*a

#### Attachments

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gneill said:

As you say, the acceleration g is negative as is the net acceleration a. Tension force is upwards. So:

T - M*g = -M*a

I see how you set that up. Thank you. I think I end up with the same answer as you do if I follow that, if I'm not mistaken. Is the answer in my edit correct? And by answer, I mean the number result, not how I got there. You showed me the correct equation set up.

iJamJL said:
I see how you set that up. Thank you. I think I end up with the same answer as you do if I follow that, if I'm not mistaken. Is the answer in my edit correct? And by answer, I mean the number result, not how I got there. You showed me the correct equation set up.

Your tension force, T = -53.827 is not correct. Also, tension force is acting upwards (positive) on the block.

gneill said:
Your tension force, T = -53.827 is not correct. Also, tension force is acting upwards (positive) on the block.

Woops. I don't know how I missed that, lol.

T-mg= -ma
T= mg-ma
T=m(g-a)=40.9*(9.81-3.09)
T=274.848N
W=274.848*8.01=2201.5 J

iJamJL said:
Woops. I don't know how I missed that, lol.

T-mg= -ma
T= mg-ma
T=m(g-a)=40.9*(9.81-3.09)
T=274.848N
W=274.848*8.01=2201.5 J

That's better!

You might want to make sure that your significant figures match those of the given information

gneill said:
That's better!

You might want to make sure that your significant figures match those of the given information

The answer came out wrong according to my online homework. Normally it requires three or more significant digits, and so I just entered the full decimal. It gave it to me as incorrect.

iJamJL said:
The answer came out wrong according to my online homework. Normally it requires three or more significant digits, and so I just entered the full decimal. It gave it to me as incorrect.

It also occurs to me that since the distance traveled is negative, and the tension force (on the block) is positive, technically the work done by the tension will be negative... work is actually being done on the rope by the block.

## What is Work by Tension for Mass Moving Down w/ Constant Acc.: WT?

Work by Tension for Mass Moving Down w/ Constant Acc.: WT is a concept in physics that explains the amount of work done by tension when a mass is moving down with a constant acceleration. It is a crucial concept in understanding the relationship between force, mass, and acceleration.

## What factors affect the work done by tension in this scenario?

The work done by tension is affected by the magnitude of the force applied, the mass of the object, and the acceleration of the object. These factors determine the overall amount of work done by tension.

## How is the work by tension calculated in this scenario?

The work done by tension is calculated by multiplying the magnitude of the tension force by the displacement of the object in the direction of the force. This can be represented by the equation WT = T * d.

## Is the work by tension positive or negative in this scenario?

The work by tension can be either positive or negative, depending on the direction of the displacement and the direction of the tension force. If the tension force and the displacement are in the same direction, the work by tension is positive. If they are in opposite directions, the work by tension is negative.

## What is the significance of understanding work by tension in this scenario?

Understanding work by tension is essential in various real-life applications, such as pulley systems, elevators, and ziplines. It also helps in understanding the relationship between force, mass, and acceleration in a more in-depth and practical way.

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