How to Calculate Work Done by Tension in a Vertical Rope for a Moving Mass

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Homework Help Overview

The discussion revolves around calculating the work done by tension in a vertical rope while lowering a block with a specified mass and acceleration. The problem involves concepts from mechanics, particularly relating to forces and work done in a gravitational field.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between force, acceleration, and work done, with some attempting to apply the equation W = m(g - a)d. Questions arise regarding the correct application of the definition of work and the use of vector notation in the context of the problem.

Discussion Status

The discussion includes various interpretations of the work done equation, with some participants expressing uncertainty about the correctness of their approaches. There is acknowledgment of the definitions involved, but no consensus has been reached on the solution. One participant claims to have found a solution, though it is unclear if this aligns with the group's understanding.

Contextual Notes

Participants are navigating the complexities of applying Newton's laws and the definition of work, with some expressing confusion over the mathematical representation of their reasoning. The problem context includes specific values for mass, acceleration, and distance, which are critical to the discussion.

NasuSama
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Sorry for the second post, but I want to know how to solve such problem when it comes to this situation.

Homework Statement



A vertical rope is used to lower a block of mass M = 48.3 kg at a constant acceleration of magnitude a = 1.54 m/s2. Find WT, the work done by the tension in the cord if the mass moves down distance s = 2.47 m.

Homework Equations



→W = mad

The Attempt at a Solution



Hm... Tried to use this form:

W = m(g - a)d

But I'm not sure if I'm at the right path.
 
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It seems correct to me :)
m(g-a) is the force used to reduce the block acceleration.
then m(g-a)d is the work done..nothing else to be considered anymore from here
 
Hi NasuSama! :smile:
NasuSama said:
→W = mad

No.

The definition of work done is W = F "dot" d
Hm... Tried to use this form:

W = m(g - a)d

But I'm not sure if I'm at the right path.

How did you get that equation? Did you use F = ma?

If so, your F is correct, your d is correct, but your "dot" isn't. :wink:
 
tiny-tim said:
Hi NasuSama! :smile:No.

The definition of work done is W = F "dot" dHow did you get that equation? Did you use F = ma?

If so, your F is correct, your d is correct, but your "dot" isn't. :wink:

So W = F * d [Can't do the dot].

I used F = ma to get the equation. What about the answer? Is it correct. You said F is correct and d is correct, right?

My answer is around 987 J, but it's incorrect.
 
W = m(g - a) [dot] d doesn't seem quite right.
 
Never mind. I got the solution!
 

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