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

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SUMMARY

The discussion focuses on calculating the work done by tension (WT) when lowering a block of mass M = 40.9 kg at a constant acceleration of a = 3.09 m/s² over a distance of s = 8.01 m. The correct formula for tension is derived as T = m(g - a), leading to T = 274.848 N. Consequently, the work done by tension is calculated as W = T * s, resulting in WT = 2201.5 J. The importance of considering significant figures and the direction of forces in the context of work is also emphasized.

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  • Basic concepts of tension in physics
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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.


Homework Equations


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.
 
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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:
attachment.php?attachmentid=44646&stc=1&d=1330716925.gif


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

T - M*g = -M*a
 

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gneill said:
attachment.php?attachmentid=44646&stc=1&d=1330716925.gif


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! :smile:

You might want to make sure that your significant figures match those of the given information :wink:
 
gneill said:
That's better! :smile:

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

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 incorrect.
 
  • #10
iJamJL said:
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 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.
 

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